poj 2386 Lake Counting 水水的搜索模板

题目啊：

Lake Counting

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 34736 Accepted: 17247

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

贴广搜代码不解释=.=

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int map[110][110], x[9]={0,0,1,0,-1,1,-1,1,-1},
team[10000][3],y[9]={0,1,0,-1,0,1,1,-1,-1},
n,m,ans;
bool map1[110][110];
char a;

void bfs(int a,int b)
{
memset(team,0,sizeof(team));
int h=0,t=1;
team[1][1]=a,team[1][2]=b,map1[a][b]=1;
do
{
h++;
for(int i=1;i<=8;i++)
{
int xx,yy;
xx=team[h][1]+x[i];
yy=team[h][2]+y[i];
if(map[xx][yy]&&!map1[xx][yy])
{
map1[xx][yy]=1;
map [xx][yy]=0;
t++;
team[t][1]=xx;
team[t][2]=yy;
}
}
}while(h<t);
}

int main()
{
//  freopen("in.txt","r",stdin);
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
getchar();
for(int j=1;j<=m;j++)
{
scanf("%c",&a);
if(a=='W')
map[i][j]=1;
else map[i][j]=0;
}
}

for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(map[i][j])
{
bfs(i,j);
ans++;
}
}

printf("%d",ans);
}

果然要时不时练练手=.=||