poj 2386 Lake Counting

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25183 Accepted: 12688

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

一道典型的dfs，思路并不难，，可是在计算一个点周围的八个点的时候，，我是单独一个一个写出来的，，

后来看了挑战程序设计，才知道大牛用的两层循环解决，，而且是是把遍历后的'W'改成'.'而不是像我一样另外设置一个flag数组（既麻烦又容易错），大牛，，这种小地方在比赛中时往往可以拉开差距

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[101][101];int num,n,m;bool bian(int x,int y){    map[x][y]='.';    for(int i=-1;i<=1;i++)        for(int j=-1;j<=1;j++)          if(map[x+i][y+j]=='W')             bian(x+i,y+j);    return 0;}int main(){    while(cin>>n>>m)    {        num=0;        memset(map,0,sizeof(map));        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)             cin>>map[i][j];        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)            if(map[i][j]=='W')             {                 bian(i,j);                 num++;             }        cout<<num<<endl;    }    return 0;}