POJ 2386 Lake Counting

Lake Counting

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.

代码

#include<cstdio>

using namespace std;

int N,M;
char filed[1005][1005];

void dfs(int x,int y){
    
    filed[x][y]='.';
    for(int a=-1;a<=1;a++)      //循环遍历八个方向
        for(int b=-1;b<=1;b++){ 
            int nx=x+a,ny=y+b;
            if(nx<N && nx>=0 && ny>=0 && ny<M && filed[nx][ny]=='W')
                dfs(nx,ny);
        }
    return ;
}

int main(){
    int arr=0;
    scanf("%d %d",&N,&M);
    for(int i=0;i<N;i++)
        scanf("%s",filed[i]);
    for(int i=0;i<N;i++)
        for(int j=0;j<M;j++){
            if(filed[i][j]=='W'){
                dfs(i,j);
                arr++;      //DP次数即答案
            }
        }
    printf("%d\n",arr);
}

深搜从最开始状态出发，遍历所有可以到达的状态。