【POJ 2386】 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

Sample Output

3

解题思路

给定一张有积水的图，判断形成几个水塘。
每个“w”是8连通的，连在一起的“w”算一个pond。可以用DFS来搜索与当前的“W”连通的所有“W”，并把搜索过的做标记，防止二次搜索。
最终搜索的次数就是形成的pond的个数。

参考代码

#include <stdio.h>const int maxn = 110;char field[maxn][maxn]={0};int n,m;void DFS(int x,int y){    field[x][y] = '.';//搜索过的做标记    for (int dx = -1;dx <= 1;dx++)        for (int dy = -1;dy <= 1;dy++){//八方向搜索            int nx = x +dx,ny = y +dy;            if (field[nx][ny] == 'W')                DFS(nx,ny);        }}int main(){    scanf("%d%d",&n,&m);    getchar();    for (int i = 1;i <= n;i++){        for (int j = 1;j <= m;j++)            field[i][j] = getchar();        getchar();    }    int ans = 0;    for (int i = 1;i <= n;i++)        for (int j = 1;j <= m;j++)            if (field[i][j] == 'W'){                DFS(i,j);                ans ++;            }    printf("%d\n",ans);    return 0;}