poj 2386 BFS+枚举

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在受近期洪水影响的农民约翰的田地中，积水形成了多个水池。通过输入田地尺寸和积水区域图，该算法计算并确定了田地中形成的水池总数。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24280		Accepted: 12255

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

Hint

OUTPUT DETAILS:

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

Source

USACO 2004 November

稍改下用DFS也行,求有几部分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<queue>
#define INF 0x3f3f3f3f

using namespace std;

struct node
{  int x;   int y;
} p;
queue<node>Q;
int n,m;
char a[200][200];
int dx[]= {-1,0,1,0,-1,-1,1,1};
int dy[]= {0,-1,0,1,1,-1,-1,1};
int panduan(int x,int y)
{
    if(x<n&&y<m&&x>=0&&y>=0&&a[x][y]=='W')
        return 1;
    return 0;
}
int BFS(int x,int y)
{
    if(!panduan(x,y))
    return 0;
    a[x][y]='*';
    Q.push(p);
    while(!Q.empty())
    {
        node q=Q.front();
        a[q.x][q.y]='*';
        Q.pop();
        for(int i=0; i<8; i++)
        {
            int fx=dx[i]+q.x;
            int fy=dy[i]+q.y;
            if(panduan(fx,fy))
               {
                   p.x=fx;p.y=fy;
                   Q.push(p);
               }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(!n&&!m)  break;
        for(int i=0; i<n; i++)
        {
            scanf("%s",a[i]);
        }
        int num=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                if(a[i][j]=='W')//看有几个连续的部分
                {
                    p.x=i;   p.y=j;
                    BFS(i,j);
                    num++;
                }
            }
        printf("%d\n",num);
    }
}