http://acm.timus.ru/problem.aspx?space=1&num=1923 -- timus

This is s problem about thd dfs and searching like the leetcode islands problems

Creating three 2-D arrays: board, land(copy), visit.

board: the input, land:(0: undecided, 1: B country, 2: M country), visit

invading the land one by one.

for each fill, operation:

if land[i][j] = enemy return;

if land[i][j] = 0 && board[i][j] = alliance: land[i][j] = empire // 1 or 2

if land[i][j] = empire; check the neighbors of each location from board.

Heads-up: checking the boundary and visited(stack overflow)

import java.util.Scanner;

public class timus1923 {
static int visit[][] = new int[55][55];
//empire B: 1, M:2, a: original land, copy: terriority
static void fill(int x, int y, int alligance, int empire,int a[][], int copy[][],int n, int m){
if(x<0 || y<0 || x>=n || y>=m) return;//our of boundary
//check the empire board
if(visit[x][y] == 1) return;
visit[x][y] = 1;
int enemy = 0;
if(empire==1) enemy = 2;
else enemy =1; if(copy[x][y] == enemy) return;//enemy's land
if(copy[x][y]==0 && a[x][y] == alligance) copy[x][y] = empire;
if(copy[x][y]==empire) {
fill(x-1,y,alligance,empire,a,copy,n,m);
fill(x,y-1,alligance,empire,a,copy,n,m);
fill(x+1,y,alligance,empire,a,copy,n,m);
fill(x,y+1,alligance,empire,a,copy,n,m);
} }
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int a[][] = new int[n+1][m+1];
int copy[][] = new int[n+1][m+1];// emprire land
//int tag[] = new int[5]; for(int i = 0; i<n; i++){
String str = in.next();
for(int j = 0; j< m; j++){
a[i][j] = str.charAt(j)-'0';
//tag[a[i][j]] ++;
}
}
int l = in.nextInt();
int b[] = new int[l+1];
for(int i = 0; i<l; i++){
b[i] = in.nextInt();
}
int countB = 0;
int countM = 0;
copy[n-1][0] = 1;
copy[0][m-1] = 2;
//start from
fill(n-1,0,a[n-1][0],1,a,copy,n,m);
fill(0,m-1,a[0][m-1],2,a,copy,n,m); for(int i = 0; i<l; i++){
//init visi
for(int j = 0; j<55; j++){
for(int k = 0; k<55; k++){
visit[j][k] = 0;
}
} if(i%2==0){
fill(n-1,0,b[i],1,a,copy,n,m);
}else{
fill(0,m-1,b[i],2,a,copy,n,m);
}
}
for(int i =0; i<n; i++){
for(int j = 0; j<m; j++){
if(copy[i][j] == 1) countB++;
else if(copy[i][j] == 2) countM++;
}
} System.out.println(countB);
System.out.println(countM);
}
}

Life is funny.

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