poj-2386 lake counting（搜索题）

Time limit1000 ms

Memory limit65536 kB

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.

 

题意：W是水洼，连起来的W算同一个水洼（是九宫格内的连起来）

题解：dfs搜索，搜索不到了就继续，每一个dfs都可以搜到一个水坑，简而言之，总的dfs的次数就是水坑的个数（dfs重新调用的dfs不算）

#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define PI 3.14159265358979323846264338327950

int N,M;
const int MAX_N=;
char field[MAX_N][MAX_N];

void dfs(int x,int y)
{
    field[x][y]='.';
    for(int dx=-;dx<=;dx++)
    {
        for(int dy=-;dy<=;dy++)
        {
            int nx=x+dx,ny=y+dy;
            if(<=nx && nx<N && <=ny && ny<M && field[nx][ny]=='W')
                dfs(nx,ny);
        }
    }
    return ;
}

void solve()
{
    int res =;
    for(int i=;i<N;i++)
    {
        for(int j=;j<M;j++)
        {
            if(field[i][j]=='W')
            {
                dfs(i,j);
                res++;
            }
        }
    }
    printf("%d\n",res);
}

int main()
{
    cin>>N>>M;
    for(int i=;i<N;i++)
        for(int j=;j<M;j++)
            cin>>field[i][j];
    solve();

}