poj 2386 Lake Counting 简单深搜

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

Hint

OUTPUT DETAILS:

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

大意：有一个大小为N*M的园子，雨后积了很多水。八连通的积水被认为是在一起的。请求出园子里共有多少个水洼？

#include<cstdio>#include<cstring>#include<iostream>using namespace std;char a[101][101];int s,t;void ss(int x,int y){     if(a[x][y]=='.'|| x<0||x>=s||y<0||y>=t)        return;    a[x][y]='.';    ss(x-1,y-1);    ss(x-1,y);    ss(x-1,y+1);    ss(x,y-1);    ss(x,y+1);    ss(x+1,y-1);    ss(x+1,y);    ss(x+1,y+1);}int main(){    while(cin>>s>>t)    {        for(int i=0; i<s; i++)            for(int j=0; j<t; j++)            {                cin>>a[i][j];            }        int sum=0;        for(int i=0; i<s; i++)            for(int j=0; j<t; j++)            {                if(a[i][j]=='W')                {                    sum++;                    ss(i,j);                }            }        cout<<sum<<endl;    }    return 0;}