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POJ 2386 Lake Counting

2024-07-11 17:49:45   222人阅读

Lake Counting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 19591 Accepted: 9848

Description

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.

Input

* 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.

Output

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

Sample Input

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

Sample Output

3


搜索水题，不解释！！！！！！！！

AC代码如下：

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

int x[8]={0,1,1,1,0,-1,-1,-1};
int y[8]={-1,-1,0,1,1,1,0,-1};
int n,m;
char b[105][105];

void sreach(int h,int z)
{
    int i;
        b[h][z]='.';
        for(i=0;i<8;i++)
        {
            if(b[h+x[i]][z+y[i]]=='W')
                sreach(h+x[i],z+y[i]);
        }
}

int main()
{

    int i,j,cont;
    while(~scanf("%d %d",&n,&m))
    {
        memset(b,0,sizeof(b));
        cont=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",b[i]);
        }
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                if(b[i][j]=='W')
                {cont++;sreach(i,j);}
            }
            printf("%d\n",cont);
    }
    return 0;
}


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