POJ 2386 Lake Counting

Lake Counting

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 22854 Accepted: 11526

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 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

深搜求 有多少个连通块，简单。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int vis[105][105];
char map[105][105];
int sum;
int n,m;
int x[5]={-1,0,1};
int y[5]={-1,0,1};
void dfs(int h,int w)
{
// if(h>n||w>m||h<1||w<1) return ;
map[h][w]='.';
for(int i=0;i<3;i++)
{
for(int j=0;j<3;j++)
{
if(h+x[i]>0&&h+x[i]<=n&&w+y[j]>0&&w+y[j]<=m&&map[h+x[i]][w+y[j]]=='W')
dfs(h+x[i],w+y[j]);
}
}
}
int main(void)
{
// freopen("B.txt","r",stdin);
// int  n,m;
scanf("%d%d",&n,&m);
sum=0;
// memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
getchar();
for(int j=1;j<=m;j++)
{
scanf("%c",&map[i][j]);
// if(map[i][j]=='.')
// vis[i][j]=1;
}
}
// dfs(1,1);
/* for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
printf("%c",map[i][j]);
}
printf("\n");
}*/
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(map[i][j]=='W')
{
// printf("haha\n");
sum++;
dfs(i,j);
}
}
}
printf("%d\n",sum);
return 0;
}