2386:Lake Counting-poj

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

输入

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

输出

* Line 1: The number of ponds in Farmer John's field.

样例输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出

3

提示

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

代码：递归 搜索

#include<stdio.h>

int n,m;
char map[110][110];
int go[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};//方向数组，上下左右--八个方向
void dfs(int x,int y)
{
        map[x][y]='.';//标记已经走过的节点
    for(int i=1;i<=8;i++)
    {
        int nx=x+go[i-1][0];//搜索上下左右斜边八个方向
        int ny=y+go[i-1][1];
        if(nx>=0&&nx<n&&ny>=0&&ny<m&&map[nx][ny]=='W')//遇到w并且没有越界  就一直遍历下去
            dfs(nx,ny);
    }

}

int main(void)
{
    int i,j,k;
    while(scanf("%d%d",&n,&m)==2)
    {
        getchar();
        k=0;
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                scanf("%c",&map[i][j]);
            }
            getchar();
        }
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                if(map[i][j]=='W')
                {
                    dfs(i,j);
                    k++;//每次找到一个水域，计数值增加1
                }
                printf("%d\n",k);
    }
    return 0;
}