HDU 1312.Red and Black-DFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19690    Accepted Submission(s): 11965

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

 

Sample Input

Sample Output

45

59

6

13

 

 

 

 

代码:

#include<bits/stdc++.h>
using namespace std;
int direct [4][2]={-1,0,1,0,0,1,0,-1};
char str[25][25];
bool flag[25][25];
int w,h,ans;
void DFS(int x,int y){
    for(int i=0;i<4;i++){
        int p=x+direct[i][0];
        int q=y+direct[i][1];
        if(p>=0&&p<h&&q>=0&&q<w&&flag[p][q]==0&&str[p][q]=='.'){
            ans++;
            flag[p][q]=1;
            DFS(p,q);
        }
    }
}
int main(){
    int Dx,Dy;
    while(~scanf("%d%d",&w,&h)){
        if(w==0&&h==0)break;
        memset(flag,0,sizeof(flag));
        getchar();
        for(int i=0;i<h;i++){
            for(int j=0;j<w;j++){
                scanf("%c",&str[i][j]);
                if(str[i][j]=='@'){
                    Dx=i;
                    Dy=j;
                }
            }getchar();
        }
        ans=1;
        flag[Dx][Dy]=1;
        DFS(Dx,Dy);
        printf("%d
",ans);
    }
    return 0;
}