标签：

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11531    Accepted Submission(s): 7177

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

45
59
6
13

#include <stdio.h>
#include <string.h>

int n,m,cnt;
char map[30][30];
int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};

void dfs(int i,int j)
{
	int k,x,y; 
    cnt++;
    map[i][j] = '#';
    for(k = 0; k<4; k++)
    {
        x = i + to[k][0];
        y = j + to[k][1];
        if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')//控制 x，y不让它越界 
            dfs(x,y);
    }
}

int main()
{
    int i,j,fi,fj;
    while(scanf("%d%d%",&m,&n))//横向为 n，纵向为 m 
    {
    	if(m==0&&n==0)	break;
        for(i = 0; i<n; i++)
        {
            for(j = 0; j<m; j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j] == '@')//找到刚开始人所在的位置 
                {
                    fi = i;
                    fj = j;
                }
            }
            getchar();
        }
        cnt = 0;
        dfs(fi,fj);
        printf("%d\n",cnt);
    }

    return 0;
}

HDU 1312 Red and Black【深搜】

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原文地址：http://blog.csdn.net/qq_16767427/article/details/44873233