[POJ] 2386 Lake Counting

Lake Counting
Time Limit: 1000MS      Memory Limit: 65536K
Total Submissions: 38083        Accepted: 18919
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
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

联通块，强迫自己用一用bfs。

//Writer:GhostCai && His Yellow Duck

#include<iostream>
#include<queue>
#include<cstdio>
#define MAXN 200
using namespace std;

int m,n;
char a[MAXN][MAXN];
int ans;
bool vis[MAXN][MAXN];

int dx[8]={0,1,1,1,0,-1,-1,-1};
int dy[8]={1,1,0,-1,-1,-1,0,1};

struct point{
    int x,y; 
}node,r;

void bfs(int x,int y){
    queue<point> Q;
    node.x = x;
    node.y = y;
    Q.push(node);
    while(!Q.empty() ){
        r=Q.front() ;
        Q.pop() ;
        for(int i=0;i<=7;i++){
            int nx=r.x+dx[i],ny=r.y+dy[i];
            if(nx<1||nx>m||ny<1||ny>n) continue;
            if(!vis[nx][ny]&&a[nx][ny]=='W'){
                vis[nx][ny]=1;
                a[nx][ny]='.';
                node.x = nx;
                node.y = ny;
                Q.push(node); 
            }
        }
    } 
}



int main(){
    cin>>m>>n;
    int i,j;
    for(i=1;i<=m;i++){
        scanf("%s",a[i]+1); 
    }
    for(i=1;i<=m;i++){
        for(j=1;j<=n;j++){
            if(!vis[i][j]&&a[i][j]=='W'){
                a[i][j]='.';
                bfs(i,j);
                ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0; 
}