Saturday, February 28, 2015

Lazy propagation in Segment Trees (SPOJ : HORRIBLE)

Lazy propagation in Segment Trees.

Read this first (the lazy propagation code): http://se7so.blogspot.in/2012/12/segment-trees-and-lazy-propagation.html

The code in the link above is for problems involving finding max in a given range.

To solve http://www.spoj.com/problems/HORRIBLE/ we need a little bit of modification in the code.

The trick is to add the new value but before that, multiplying it by the number of nodes below it during lazy updation at any node.

Code (C++):

#include <stdio.h>
#include <math.h> 

long long *segmentTree;
long long *lazy;

long long query(int qLo, int qHi, int cLo, int cHi, int stIndex);

void update(int qLo, int qHi, int cLo, int cHi, int stIndex, long long val) {
    if (cHi < qLo || qHi < cLo)return;

    if (lazy[stIndex] != 0) {
        if (cLo == cHi) {
            segmentTree[stIndex] += lazy[stIndex];
        } else {
            segmentTree[stIndex] += lazy[stIndex]*(cHi - cLo + 1);
            lazy[stIndex * 2 + 1] += lazy[stIndex];
            lazy[stIndex * 2 + 2] += lazy[stIndex];
        }
        lazy[stIndex] = 0;
    }

    if (qLo <= cLo && cHi <= qHi) {
        if (cLo == cHi) {
            segmentTree[stIndex] += val;
        } else {
            segmentTree[stIndex] += val * (cHi - cLo + 1);
            lazy[stIndex * 2 + 1] += val;
            lazy[stIndex * 2 + 2] += val;
        }
        return;
    }

    int mid = (cLo + cHi) / 2;
    update(qLo, qHi, cLo, mid, stIndex * 2 + 1, val);
    update(qLo, qHi, mid + 1, cHi, stIndex * 2 + 2, val);
    int left = stIndex * 2 + 1;
    int right = left + 1;
    segmentTree[stIndex] = query(cLo, mid, cLo, mid, stIndex * 2 + 1) + query(mid + 1, cHi, mid + 1, cHi, stIndex * 2 + 2);
}

long long query(int qLo, int qHi, int cLo, int cHi, int stIndex) {
    if (cHi < qLo || qHi < cLo)return 0;

    if (lazy[stIndex] != 0) {
        if (cLo == cHi) {
            segmentTree[stIndex] += lazy[stIndex];
        } else {
            segmentTree[stIndex] += lazy[stIndex]*(cHi - cLo + 1);
            lazy[stIndex * 2 + 1] += lazy[stIndex];
            lazy[stIndex * 2 + 2] += lazy[stIndex];
        }
        lazy[stIndex] = 0;
    }
    if (qLo <= cLo && cHi <= qHi)return segmentTree[stIndex];

    int mid = (cLo + cHi) / 2;
    long long left = query(qLo, qHi, cLo, mid, stIndex * 2 + 1);
    long long right = query(qLo, qHi, mid + 1, cHi, stIndex * 2 + 2);
    return left + right;
}

int segSize;

void constructSegmentTree(int size) {
    int height = (int) ceil(log((double) size) / log((double) 2));
    segSize = (int) ceil(pow((double) 2, height + 1));
    segmentTree = new long long[segSize];
    lazy = new long long[segSize];
    for (int i = 0; i < segSize; ++i) {
        segmentTree[i] = 0;
        lazy[i] = 0;
    }
}

int main() {
    int nCases;
    scanf("%d", &nCases);
    while (nCases--) {
        int N, C;
        scanf("%d %d", &N, &C);
        constructSegmentTree(N);
        for (int i = 0; i < C; ++i) {
            int type, p, q;
            long long v;
            scanf("%d", &type);
            if (type == 0) {
                scanf("%d %d %lld", &p, &q, &v);
                update(p - 1, q - 1, 0, N - 1, 0, v);
            } else {
                scanf("%d %d", &p, &q);
                printf("%lld\n", query(p - 1, q - 1, 0, N - 1, 0));
            }
        }
        delete(segmentTree);
        delete(lazy);
    }
}


Thursday, February 26, 2015

Segment Trees : SPOJ.com problem GSS1 & GSS3 solution.

Segment Trees Tutorials: 


C++ Code  - GSS1:

#include <stdio.h>
#include <math.h>

class Node {
public:
    int lMax;
    int rMax;
    int anyMax;
    int sum;

    Node(int l, int a, int r, int s) {
        lMax = l;
        rMax = r;
        anyMax = a;
        sum = s;
    }
};

Node **segmentTree;
int *array;

int max(int a, int b) {
    if (a > b)return a;
    else return b;
}

void segmentTreeConstructor(int arrayLo, int arrayHi, int stIndex) {
    if (arrayLo == arrayHi) {
        segmentTree[stIndex] = new Node(array[arrayLo], array[arrayLo], array[arrayLo], array[arrayLo]);
        return;
    }

    int mid = (arrayLo + arrayHi) / 2;
    segmentTreeConstructor(arrayLo, mid, stIndex * 2 + 1);
    segmentTreeConstructor(mid + 1, arrayHi, stIndex * 2 + 2);

    Node *current = segmentTree[stIndex] = new Node(0, 0, 0, 0);
    Node *leftChild = segmentTree[stIndex * 2 + 1];
    Node *rightChild = segmentTree[stIndex * 2 + 2];

    current->lMax = max(leftChild->lMax, leftChild->sum + rightChild->lMax);
    current->rMax = max(leftChild->rMax + rightChild->sum, rightChild->rMax);
    current->anyMax = max(leftChild->anyMax, max(leftChild->rMax + rightChild->lMax, rightChild->anyMax));
    current->sum = leftChild->sum + rightChild->sum;
}

Node& find(int stIndex, int qLow, int qHi, int currentLo, int currentHi) {
    if (qLow == currentLo && qHi == currentHi)return *(segmentTree[stIndex]);
    int mid = (currentLo + currentHi) / 2;
    if (qHi <= mid)return find(2 * stIndex + 1, qLow, qHi, currentLo, mid);
    else if (qLow > mid)return find(2 * stIndex + 2, qLow, qHi, mid + 1, currentHi);
    else {
        Node *currentResult = new Node(0, 0, 0, 0);
        Node leftResult = find(2 * stIndex + 1, qLow, mid, currentLo, mid);
        Node rightResult = find(2 * stIndex + 2, mid + 1, qHi, mid + 1, currentHi);

        currentResult->lMax = max(leftResult.lMax, leftResult.sum + rightResult.lMax);
        currentResult->rMax = max(leftResult.rMax + rightResult.sum, rightResult.rMax);
        currentResult->anyMax = max(leftResult.anyMax, max(leftResult.rMax + rightResult.lMax, rightResult.anyMax));
        currentResult->sum = leftResult.sum + rightResult.sum;

        return *currentResult;
    }
}

void constructSegmentTree(int* array, int size) {
    int height = (int) ceil(log((double) size) / log((double) 2));
    segmentTree = new Node*[(int) ceil(pow((double) 2, height + 1))];
    segmentTreeConstructor(0, size - 1, 0);
}

int main() {
    int nElements;
    scanf("%d", &nElements);
    array = new int[nElements];
    for (int i = 0; i < nElements; ++i)scanf("%d", &array[i]);
    constructSegmentTree(array, nElements);
    int nQueries;
    scanf("%d", &nQueries);
    while (nQueries--) {
        int l, h;
        scanf("%d %d", &l, &h);
        Node result = find(0, l - 1, h - 1, 0, nElements - 1);
        printf("%d\n", result.anyMax);
    }
}


C++ GSS3 Code:

#include <stdio.h>
#include <math.h>

class Node {
public:
    int lMax;
    int rMax;
    int anyMax;
    int sum;

    Node(int l, int a, int r, int s) {
        lMax = l;
        rMax = r;
        anyMax = a;
        sum = s;
    }

    void toString() {
        printf("[%d %d %d %d]\n", lMax, rMax, anyMax, sum);
    }
};

Node **segmentTree;
int *array;

int max(int a, int b) {
    if (a > b)return a;
    else return b;
}

void segmentTreeConstructor(int arrayLo, int arrayHi, int stIndex) {
    if (arrayLo == arrayHi) {
        segmentTree[stIndex] = new Node(array[arrayLo], array[arrayLo], array[arrayLo], array[arrayLo]);
        return;
    }

    int mid = (arrayLo + arrayHi) / 2;
    segmentTreeConstructor(arrayLo, mid, stIndex * 2 + 1);
    segmentTreeConstructor(mid + 1, arrayHi, stIndex * 2 + 2);

    Node *current = segmentTree[stIndex] = new Node(0, 0, 0, 0);
    Node *leftChild = segmentTree[stIndex * 2 + 1];
    Node *rightChild = segmentTree[stIndex * 2 + 2];

    current->lMax = max(leftChild->lMax, leftChild->sum + rightChild->lMax);
    current->rMax = max(leftChild->rMax + rightChild->sum, rightChild->rMax);
    current->anyMax = max(leftChild->anyMax, max(leftChild->rMax + rightChild->lMax, rightChild->anyMax));
    current->sum = leftChild->sum + rightChild->sum;
}

Node& find(int stIndex, int qLow, int qHi, int currentLo, int currentHi) {
    if (qLow == currentLo && qHi == currentHi)return *(segmentTree[stIndex]);
    int mid = (currentLo + currentHi) / 2;
    if (qHi <= mid)return find(2 * stIndex + 1, qLow, qHi, currentLo, mid);
    else if (qLow > mid)return find(2 * stIndex + 2, qLow, qHi, mid + 1, currentHi);
    else {
        Node *currentResult = new Node(0, 0, 0, 0);
        Node leftResult = find(2 * stIndex + 1, qLow, mid, currentLo, mid);
        Node rightResult = find(2 * stIndex + 2, mid + 1, qHi, mid + 1, currentHi);

        currentResult->lMax = max(leftResult.lMax, leftResult.sum + rightResult.lMax);
        currentResult->rMax = max(leftResult.rMax + rightResult.sum, rightResult.rMax);
        currentResult->anyMax = max(leftResult.anyMax, max(leftResult.rMax + rightResult.lMax, rightResult.anyMax));
        currentResult->sum = leftResult.sum + rightResult.sum;

        return *currentResult;
    }
}

void modify(int stIndex, int target, int currentLo, int currentHi, int newVal) {
    if (target == currentLo && currentLo == currentHi) {
        Node &target = *(segmentTree[stIndex]);
        target.sum = newVal;
        target.lMax = newVal;
        target.rMax = newVal;
        target.anyMax = newVal;
        return;
    }

    int mid = (currentLo + currentHi) / 2;
    if (target <= mid) modify(2 * stIndex + 1, target, currentLo, mid, newVal);
    else if (target > mid) modify(2 * stIndex + 2, target, mid + 1, currentHi, newVal);

    Node *currentResult = segmentTree[stIndex];

    Node leftResult = *(segmentTree[2 * stIndex + 1]);
    Node rightResult = *(segmentTree[2 * stIndex + 2]);

    currentResult->lMax = max(leftResult.lMax, leftResult.sum + rightResult.lMax);
    currentResult->rMax = max(leftResult.rMax + rightResult.sum, rightResult.rMax);
    currentResult->anyMax = max(leftResult.anyMax, max(leftResult.rMax + rightResult.lMax, rightResult.anyMax));
    currentResult->sum = leftResult.sum + rightResult.sum;
}


void constructSegmentTree(int* array, int size) {
    int height = (int) ceil(log((double) size) / log((double) 2));
    segmentTree = new Node*[(int) ceil(pow((double) 2, height + 1))];
    n = (int) ceil(pow((double) 2, height + 1));
    segmentTreeConstructor(0, size - 1, 0);
}

int main() {
    int nElements;
    scanf("%d", &nElements);
    array = new int[nElements];
    for (int i = 0; i < nElements; ++i)scanf("%d", &array[i]);
    constructSegmentTree(array, nElements);
    int nQueries;
    scanf("%d", &nQueries);
    while (nQueries--) {
        int type, l, h;
        scanf("%d %d %d", &type, &l, &h);
        if (type == 1) {
            Node result = find(0, l - 1, h - 1, 0, nElements - 1);
            printf("%d\n", result.anyMax);
        } else {
            modify(0, l - 1, 0, nElements - 1, h);
        }
    }
}

Thursday, February 19, 2015

Checking if a given graph is Bipartite (BFS) JAVA

Algorithm:

1. Use a queue q.
2. Add any node to q.
3. For all those nodes that are connected to the current node polled from q, check if they have been assigned a color, if they have a color and it is the same as the current node, the graph is not bipartite and we need to stop the procedure else if they don't have any color add them to queue also (only if they have not been added previously [ use a boolean array or a HashSet for checking this ] ).
4. If the queue q is empty in the end it means that the given graph is bipartite.

Note: In case the graph is disjoint, we need to add all the given nodes to the queue and then check for all components connected to the current node popped from the queue.

Source:

import java.util.*;

public class BipartiteCheck {
    public static void main(String[] args) throws Exception{ 
        //Indices start from 1
            int graph[][] = {
                {0,0,0,0,0,0,0,0,0},
                {0,0,1,0,1,0,1,0,0},
                {0,1,0,1,0,1,0,0,0},
                {0,0,1,0,1,0,0,0,0},
                {0,1,0,1,0,1,0,0,0},
                {0,0,1,0,1,0,1,0,0},
                {0,1,0,0,0,1,0,1,0},
                {0,0,0,0,0,0,1,0,1},
                {0,0,0,0,0,0,0,1,0},
            };
            
            int[] color = new int[graph.length]; 
            color[1] = 1;
            
            Queue<Integer> q = new LinkedList<>();
            
            //Can also use a HashSet
            boolean checked[] = new boolean[graph[0].length];
            
            q.add(1); 
            
            boolean check = true;
            
            //In case the graph has disjoint nodes, add all of them to queue
            //This is the case when every node of the graph is reachable from every other node
            while(!q.isEmpty()){
                int node = q.poll(); 
//                System.out.println("Checking node "+node);
                for(int i=node+1;i<graph[0].length;++i){   
                    if(graph[node][i]==1){ 
//                        System.out.println("node "+node+" is connected to node "+i);
                        if(!checked[i]){ q.add(i); checked[i] = true;}
                        if(color[i] == color[node]){
                            check = false;
                            break;
                        } 
//                        System.out.println("Assigning color = "+(-color[node])+" to node "+i);
                        color[i] = -color[node]; 
                    }
                }
                if(!check)break;
            }
            System.out.println(Arrays.toString(color));
            if(check) System.out.println("Given graph is bipartite!");
            else System.out.println("Given graph is NOT bipartite!");
    }
}

Output:

[0, 1, -1, 1, -1, 1, -1, 1, -1]
Given graph is bipartite!

Graph Coloring Problem : Backtracking Solution Java Implementation

Tutorial : https://www.youtube.com/watch?v=Cl3A_9hokjU

Source:

import java.util.*;
import java.io.*; 
import java.math.*;

public class GraphColoring {
    static int connected[][];
    static int[] colors;
    static int nColors;
    static int nNodes;
    
    static int getNodeColor(int k){
        do{ 
            int j;
            //Assign the next color
            colors[k] = colors[k]+1; 
            
            //If all colors have been tested on this node, 
            //return because there are no more new colors left 
            //to be considered for this node
            if(colors[k]==nColors+1)return 0;
            
            //Check to see if some connected node already has this color
            for(j=1;j<=nNodes;++j){ 
                if(connected[k][j] == 1 && colors[k] == colors[j] && k!=j){ 
                    break;
                }
            }  
            if(j==nNodes+1)return colors[k];
        }while(true);
    }
    
    static void mColoring(int k){
        do{ 
            //get a color for this node
            colors[k] = getNodeColor(k); 
            
            //if no more colors can be applied to this node, return
            if(colors[k] == 0)return;
            
            //if all the nodes have been assigned colors successfully, print the color assignments
            if(k==nNodes){System.out.println("Color Assignment: "+Arrays.toString(colors));
            //return /*don't quit until all possible assignments have been found*/
            }
            //consider the next node
            else mColoring(k+1);
        }while(true);
    }
    
    public static void main(String[] args) throws Exception{ 
            nColors = 3;
            connected = new int[][]{
                {0, 0, 0, 0, 0},
                {0, 1, 1, 0, 1},
                {0, 1, 1, 1, 1},
                {0, 0, 1, 1, 1},
                {0, 1, 1, 1, 1},
            }; 
            nNodes = connected.length-1;
            colors = new int[nNodes+1];
            
            mColoring(1);
    } 
}

Output:

Color Assignment: [0, 1, 2, 1, 3]
Color Assignment: [0, 1, 3, 1, 2]
Color Assignment: [0, 2, 1, 2, 3]
Color Assignment: [0, 2, 3, 2, 1]
Color Assignment: [0, 3, 1, 3, 2]
Color Assignment: [0, 3, 2, 3, 1]

Tuesday, February 17, 2015

A Sorting Algorithm

I thought of this algorithm this evening and implemented it for fun. It's not very good but anyways:

Time Complexity : O(n^2)
Space Complexity : O(n^2)

Algorithm:

We have an array of elements.

Start with first element. Store it in a stack. Iterate over remaining elements and check to see if they are greater than the last biggest found element in this stack. If it is bigger, store it in the stack and change bigger to contain this element. If it is less, skip it.

Now for the remaining elements, start with the first remaining element. Put it in a stack. Now do the same process with this stack (Iterate over the remaining elements to see if they are bigger than the current biggest element in this stack, skipping if they are smaller).

Do this until all of the elements have been put in some stack. 

It is possible to have stacks containing only one element.

Now, compare the stack tops and pop that stack which has the biggest of all stack top elements.
Put the popped element in the original array at location i (i=0 to length of the array).

Do this until all of the stacks become empty.

The array is sorted.

Consider the array:

3, 7, 7, 4, 2, -3, 0

Stacks:

                                                                                                                             
7                                                                                                                           
7                        0                                                                                              
3       4       2       -3
S1      S2      S3      S4

Now merge these stacks

Max(7, 4, 2, 0) = 7
Max(7, 4, 2, 0) = 7
Max(3, 4, 2, 0) = 4
Max(3, 2, 0)     = 3
Max(2, 0)         = 2
Max(0)             = 0
Max(-3)            = -3      

Sorted Array  = 7, 7, 4, 3, 2, 0, -3

Elements have been sorted!                                                                                                                                                                                  
Code:

import java.util.Arrays;

public class ASort {

    public static void main(String[] args) throws Exception {
        int array[] = {6, 4, 8, 9, 6, 5, -8, -9, 5, 6, -99, -5, 5, 0, 2, 3, 5,5,6,7,8,8,9,0, 0, 2, 5, 6, 9, -9, -8, -56};

        int[][] stackAr = new int[array.length][array.length];
        int[] indices = new int[array.length];
        Arrays.fill(indices, -1);
        int index = 0;

        for (int i = 0; i < array.length; ++i) {
            if (array[i] != Integer.MAX_VALUE) {
                stackAr[index][++indices[index]] = array[i];
                int greater = array[i];

                for (int j = i + 1; j < array.length; ++j) {
                    if (array[j] != Integer.MAX_VALUE && array[j] >= greater) {
                        stackAr[index][++indices[index]] = array[j];
                        greater = array[j];
                        array[j] = Integer.MAX_VALUE;
                    }
                }
                index++;
            }
        }

        index = 0;
        for (int i = 0; i < array.length; ++i) {
            int max = Integer.MIN_VALUE;
            int indexOfTarget = 0;
            for (int j = 0; j < array.length; ++j) {
                if (indices[j] != -1) {
                    if (stackAr[j][indices[j]] > max) {
                        max = stackAr[j][indices[j]];
                        indexOfTarget = j;
                    }
                }
            }
            array[i] = stackAr[indexOfTarget][indices[indexOfTarget]--];
        }

        for (int i = array.length - 1; i > -1; --i) {
            System.out.print(array[i] + ", ");
        }
        System.out.println();
    }
}
                                                                                                                                                                     
Output:

-99, -56, -9, -9, -8, -8, -5, 0, 0, 0, 2, 2, 3, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 8, 8, 8, 9, 9, 9,                                                                                                                       

Sunday, February 8, 2015

A* Shortest Path Finding Algorithm Implementation in Java

A nice Tutorial : https://www.youtube.com/watch?v=-L-WgKMFuhE

Pseudo code:

OPEN //the set of nodes to be evaluated
CLOSED //the set of nodes already evaluated
add the start node to OPEN

loop
    current = node in OPEN with the lowest f_cost
    remove current from OPEN
    add current to CLOSED

    if current is the target node //path has been found
        return

    foreach neighbour of the current node
        if neighbour is not traversable or neighbour is in CLOSED
            skip to the next neighbour

        if new path to neighbour is shorter OR neighbour is not in OPEN
            set f_cost of neighbour
            set parent of neighbour to current 
            if neighbour is not in OPEN
                add neighbour to OPEN

Source:

import java.util.*;

public class AStar {
    public static final int DIAGONAL_COST = 14;
    public static final int V_H_COST = 10;
    
    static class Cell{  
        int heuristicCost = 0; //Heuristic cost
        int finalCost = 0; //G+H
        int i, j;
        Cell parent; 
        
        Cell(int i, int j){
            this.i = i;
            this.j = j; 
        }
        
        @Override
        public String toString(){
            return "["+this.i+", "+this.j+"]";
        }
    }
    
    //Blocked cells are just null Cell values in grid
    static Cell [][] grid = new Cell[5][5];
    
    static PriorityQueue<Cell> open;
     
    static boolean closed[][];
    static int startI, startJ;
    static int endI, endJ;
            
    public static void setBlocked(int i, int j){
        grid[i][j] = null;
    }
    
    public static void setStartCell(int i, int j){
        startI = i;
        startJ = j;
    }
    
    public static void setEndCell(int i, int j){
        endI = i;
        endJ = j; 
    }
    
    static void checkAndUpdateCost(Cell current, Cell t, int cost){
        if(t == null || closed[t.i][t.j])return;
        int t_final_cost = t.heuristicCost+cost;
        
        boolean inOpen = open.contains(t);
        if(!inOpen || t_final_cost<t.finalCost){
            t.finalCost = t_final_cost;
            t.parent = current;
            if(!inOpen)open.add(t);
        }
    }
    
    public static void AStar(){ 
        
        //add the start location to open list.
        open.add(grid[startI][startJ]);
        
        Cell current;
        
        while(true){ 
            current = open.poll();
            if(current==null)break;
            closed[current.i][current.j]=true; 

            if(current.equals(grid[endI][endJ])){
                return; 
            } 

            Cell t;  
            if(current.i-1>=0){
                t = grid[current.i-1][current.j];
                checkAndUpdateCost(current, t, current.finalCost+V_H_COST); 

                if(current.j-1>=0){                      
                    t = grid[current.i-1][current.j-1];
                    checkAndUpdateCost(current, t, current.finalCost+DIAGONAL_COST); 
                }

                if(current.j+1<grid[0].length){
                    t = grid[current.i-1][current.j+1];
                    checkAndUpdateCost(current, t, current.finalCost+DIAGONAL_COST); 
                }
            } 

            if(current.j-1>=0){
                t = grid[current.i][current.j-1];
                checkAndUpdateCost(current, t, current.finalCost+V_H_COST); 
            }

            if(current.j+1<grid[0].length){
                t = grid[current.i][current.j+1];
                checkAndUpdateCost(current, t, current.finalCost+V_H_COST); 
            }

            if(current.i+1<grid.length){
                t = grid[current.i+1][current.j];
                checkAndUpdateCost(current, t, current.finalCost+V_H_COST); 

                if(current.j-1>=0){
                    t = grid[current.i+1][current.j-1];
                    checkAndUpdateCost(current, t, current.finalCost+DIAGONAL_COST); 
                }
                
                if(current.j+1<grid[0].length){
                   t = grid[current.i+1][current.j+1];
                    checkAndUpdateCost(current, t, current.finalCost+DIAGONAL_COST); 
                }  
            }
        } 
    }
    
    /*
    Params :
    tCase = test case No.
    x, y = Board's dimensions
    si, sj = start location's x and y coordinates
    ei, ej = end location's x and y coordinates
    int[][] blocked = array containing inaccessible cell coordinates
    */
    public static void test(int tCase, int x, int y, int si, int sj, int ei, int ej, int[][] blocked){
           System.out.println("\n\nTest Case #"+tCase);
            //Reset
           grid = new Cell[x][y];
           closed = new boolean[x][y];
           open = new PriorityQueue<>((Object o1, Object o2) -> {
                Cell c1 = (Cell)o1;
                Cell c2 = (Cell)o2;

                return c1.finalCost<c2.finalCost?-1:
                        c1.finalCost>c2.finalCost?1:0;
            });
           //Set start position
           setStartCell(si, sj);  //Setting to 0,0 by default. Will be useful for the UI part
           
           //Set End Location
           setEndCell(ei, ej); 
           
           for(int i=0;i<x;++i){
              for(int j=0;j<y;++j){
                  grid[i][j] = new Cell(i, j);
                  grid[i][j].heuristicCost = Math.abs(i-endI)+Math.abs(j-endJ);
//                  System.out.print(grid[i][j].heuristicCost+" ");
              }
//              System.out.println();
           }
           grid[si][sj].finalCost = 0;
           
           /*
             Set blocked cells. Simply set the cell values to null
             for blocked cells.
           */
           for(int i=0;i<blocked.length;++i){
               setBlocked(blocked[i][0], blocked[i][1]);
           }
           
           //Display initial map
           System.out.println("Grid: ");
            for(int i=0;i<x;++i){
                for(int j=0;j<y;++j){
                   if(i==si&&j==sj)System.out.print("SO  "); //Source
                   else if(i==ei && j==ej)System.out.print("DE  ");  //Destination
                   else if(grid[i][j]!=null)System.out.printf("%-3d ", 0);
                   else System.out.print("BL  "); 
                }
                System.out.println();
            } 
            System.out.println();
           
           AStar(); 
           System.out.println("\nScores for cells: ");
           for(int i=0;i<x;++i){
               for(int j=0;j<x;++j){
                   if(grid[i][j]!=null)System.out.printf("%-3d ", grid[i][j].finalCost);
                   else System.out.print("BL  ");
               }
               System.out.println();
           }
           System.out.println();
            
           if(closed[endI][endJ]){
               //Trace back the path 
                System.out.println("Path: ");
                Cell current = grid[endI][endJ];
                System.out.print(current);
                while(current.parent!=null){
                    System.out.print(" -> "+current.parent);
                    current = current.parent;
                } 
                System.out.println();
           }else System.out.println("No possible path");
    }
     
    public static void main(String[] args) throws Exception{   
        test(1, 5, 5, 0, 0, 3, 2, new int[][]{{0,4},{2,2},{3,1},{3,3}}); 
        test(2, 5, 5, 0, 0, 4, 4, new int[][]{{0,4},{2,2},{3,1},{3,3}});   
        test(3, 7, 7, 2, 1, 5, 4, new int[][]{{4,1},{4,3},{5,3},{2,3}});
        
        test(1, 5, 5, 0, 0, 4, 4, new int[][]{{3,4},{3,3},{4,3}});
    }
}

Output: 

Test Case #1
Grid: 
SO  0   0   0   BL  
0   0   0   0   0   
0   0   BL  0   0   
0   BL  DE  BL  0   
0   0   0   0   0   


Scores for cells: 
0   14  27  41  BL  
14  17  29  42  56  
27  29  BL  45  59  
39  BL  43  BL  0   
52  55  0   0   0   

Path: 
[3, 2] -> [2, 1] -> [1, 1] -> [0, 0]


Test Case #2
Grid: 
SO  0   0   0   BL  
0   0   0   0   0   
0   0   BL  0   0   
0   BL  0   BL  0   
0   0   0   0   DE  


Scores for cells: 
0   17  33  48  BL  
17  20  35  49  62  
33  35  BL  52  64  
48  BL  52  BL  67  
62  65  64  67  77  

Path: 
[4, 4] -> [3, 4] -> [2, 3] -> [1, 2] -> [1, 1] -> [0, 0]


Test Case #3
Grid: 
0   0   0   0   0   0   0   
0   0   0   0   0   0   0   
0   SO  0   BL  0   0   0   
0   0   0   0   0   0   0   
0   BL  0   BL  0   0   0   
0   0   0   BL  DE  0   0   
0   0   0   0   0   0   0   


Scores for cells: 
40  35  37  40  53  68  0   
22  17  20  34  48  63  0   
17  0   15  BL  48  61  75  
20  15  18  31  43  56  70  
34  BL  31  BL  46  58  73  
48  48  43  BL  56  61  0   
63  61  56  59  0   0   0   

Path: 
[5, 4] -> [4, 4] -> [3, 3] -> [3, 2] -> [2, 1]


Test Case #1
Grid: 
SO  0   0   0   0   
0   0   0   0   0   
0   0   0   0   0   
0   0   0   BL  BL  
0   0   0   BL  DE  


Scores for cells: 
0   17  33  48  62  
17  20  35  49  62  
33  35  38  51  63  
48  49  51  BL  BL  
62  62  63  BL  0   

No possible path 

Add a comment if you find any error :)