poj 2386 Lake Counting(BFS解法)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45142		Accepted: 22306

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.

/**********************
author: yomi
date: 18.8.14
ps:现在的行为大概像以卵击石，没办法，死也要挣扎着死。
**********************/
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;
const int maxn = 150;
bool vis[maxn][maxn];
int m, n;
char g[maxn][maxn];
struct Pos
{
    int x, y;
}p[maxn];
int dx[3] = {-1, 0, 1};
int dy[3] = {-1, 0, 1};
void bfs(int x, int y)
{
    queue<Pos>q;
    while(!q.empty()){
        q.pop();
    }
    Pos p;
    p.x = x;
    p.y = y;
    q.push(p);
    vis[x][y] = true;
    while(!q.empty()){
        Pos now = q.front();
        q.pop();
        for(int i=0; i<3; i++){
            for(int j=0; j<3; j++){
                int newX = now.x + dx[i];
                int newY = now.y + dy[j];
                if(newX<0||newX>=n||newY<0||newY>=m)
                    continue;
                if(g[newX][newY]=='.')
                    continue;
                if(!vis[newX][newY]){
                    Pos New;
                    New.x = newX;
                    New.y = newY;
                    q.push(New);
                    vis[newX][newY] = true;
                }
            }
        }
    }
}
int main()
{
    cin >> n >> m;
    int ans = 0;
    for(int i=0; i<n; i++){
        scanf("%s", g[i]);
    }
    for(int i=0; i<n; i++){
        for(int j=0; j<m; j++){
            if(g[i][j] == 'W' && !vis[i][j]){
                bfs(i, j);
                ans++;
            }
        }
    }
    cout << ans;
    return 0;
}

/**
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
**/