POJ 2386

题目：

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 21515 Accepted: 10831

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

解题思路：简单的深度优先搜索。

代码：

#include<iostream>using namespace std;#defineMAX_N 105int N,M;char field[MAX_N][MAX_N + 1]; //园子//现在位置（x,y）void dfs(int x, int y){//将现在所在位置替换为 '.'。field[x][y] = '.';//循环遍历移动的8个方向for(int dx=-1; dx<=1; dx++){for(int dy=-1; dy<=1; dy++){//向x方向移动dx， 向y方向移动dy，移动的结果为(nx,ny)int nx = x + dx, ny = y + dy;//判断(nx,ny)是不是在园子内，以及是否有积水if(0<=nx && nx<N && 0<=ny && ny<M && field[nx][ny] == 'W')dfs(nx,ny);}}return;}void solve(){int res = 0;for(int i=0; i<N; i++){for(int j=0; j<M; j++){if(field[i][j] == 'W'){//从有W的地方开始dfsdfs(i,j);res ++;}}}printf("%d\n",res);}int main(){//freopen("input.txt","r",stdin);      //freopen("output.txt","w",stdout); int i,j;while(cin>>N>>M){for(i=0; i<N; i++){for(j=0; j<M; j++){cin>>field[i][j];}}solve();}return 0;}