POJ 1979 Red and Black

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Description

Input

Output

Sample Input

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

Sample Output

45
59
6
13

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define N 1009
using namespace std;
char s[N][N];

int n,m;
int sx,sy;
int dis[4][2]={ {0,1},{0,-1},{1,0},{-1,0} };
int sum;

void dfs(int x,int y)
{
    for(int i=0;i<4;i++)
    {
        int xx=x+dis[i][0];
        int yy=y+dis[i][1];
        if(xx>=0 && xx<n && yy>=0 && yy<m && s[xx][yy]=='.')
        {
           s[xx][yy]='#';
           sum++;
           dfs(xx,yy);
        }
    }
}

int main()
{
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        if(n+m<=0) break;

        sum=1;//起点一定算一个答案,如果只有一个点的话，那么答案将输出0

        for(int i=0;i<n;i++)
        {
            scanf("%s",s[i]);
            for(int j=0;j<m;j++)
            {
                if(s[i][j]=='@')//起点已经算了一个，直接修改成#
                {
                    sx=i;
                    sy=j;
                    s[i][j]='#';
                    break;
                }
            }
        }

        dfs(sx,sy);
        printf("%d\n",sum);

    }
    return 0;
}

POJ 1979 Red and Black

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原文地址：http://blog.csdn.net/wust_zjx/article/details/45786815