POJ

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 36172 Accepted: 17965

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

Sample Output

3

题意：n*m的院子里有水（W），水只要八方有水就算连着，求院子里的水洼数量。

解决：DFS八个方向

细节：

#include<iostream>#include<algorithm>using namespace std;int n , m, num = 0;char a[110][110];void dfs(int x, int y){    a[x][y] = '.';    for(int i = -1; i <= 1; i++)    for(int j = -1; j <= 1; j++){        int nx = x + i, ny = y + j;        if(0 <= nx && 0 <= ny && nx <= n && ny <= m && a[nx][ny] == 'W')            dfs(nx, ny);    }    return;}int main(){    cin >> n >> m;    for(int i = 0; i < n; i++)        cin >> a[i];    for(int i = 0; i < n; i++)    for(int j = 0; j < m; j++){        if(a[i][j]=='W'){            dfs(i, j);            num++;        }    }    cout << num << endl;}