poj2386 Lake Counting （深搜）

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25416 Accepted: 12803

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

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

USACO 2004 November

解析：直接看代码吧。

代码：

#include<cstdio>using namespace std;const int maxn=1e2;int n,m;char s[maxn+10][maxn+10];void dye(int x,int y){  int i,j;  s[x][y]='.';  for(i=-1;i<=1;i++)    for(j=-1;j<=1;j++)      if(1<=x+i && x+i<=n &&      1<=y+j && y+j<=m && s[x+i][y+j]=='W')dye(x+i,y+j);}int main(){  freopen("1.in","r",stdin);  int i,j,ans;  while(scanf("%d%d\n",&n,&m)==2)    {      for(i=1;i<=n;i++)        for(j=1;j<=m;j++)          scanf("%c\n",&s[i][j]);      for(ans=0,i=1;i<=n;i++)        for(j=1;j<=m;j++)          if(s[i][j]=='W')dye(i,j),ans++;      printf("%d\n",ans);}   return 0;}