题目链接 http://poj.org/problem?id=2386
题目
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.
思路 这是一道经典的DFS题目
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int n,m;
char maps[105][105];
void dfs(int x,int y)
{maps[x][y]='.';for(int i=-1;i<=1;i++) // 从周围查找for(int j=-1;j<=1;j++){int newx=x+i;int newy=y+j;if(newx>=0&&newx<n&&newy>=0&&newy<m&&maps[newx][newy]=='W') // 如果符合继续迭代dfs(newx,newy);}
}
int main()
{while(cin>>n>>m){int sum=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++)cin>>maps[i][j];for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(maps[i][j]=='W'){dfs(i,j);sum++; // 每一轮迭代表示一处积水}cout<<sum<<endl;}return 0;
}