【POJ2386】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.

题意：.是土地 W是水洼    求有多少出水洼 相邻的W算是一个连通的水洼

题解：依次搜索每一个点 如果是W ans++并进入dfs 吧相邻的所有点变成.        最后输出ans

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

const int maxn = 110;
int N, M;
char fi[maxn][maxn];    //矩形区域

void dfs(int x, int y); //搜索

int main(){
    scanf("%d%d", &N, &M);
    for(int i = 0; i < N; ++i){
        scanf("%s", fi[i]);
    }

    int ans = 0;
    for(int i = 0; i < N; ++i){
        for(int j = 0; j < M; ++j){
            if(fi[i][j] == 'W'){    //如果是W进入递归
                dfs(i, j);
                ++ans;              //水洼+1
            }
        }
    }

    printf("%d\n", ans);
    return 0;
}

void dfs(int x, int y){
    fi[x][y] = '.';                 //标记为.

    for(int dx = -1; dx <= 1; ++dx){
        for(int dy = -1; dy <= 1; ++dy){
            int nx = x + dx;
            int ny = y +dy;

            if(0 <= nx && nx < N && 0 <= ny && ny < M && fi[nx][ny] == 'W'){    //搜索前后左右
                dfs(nx, ny);                                                    //如果是W进入循环
            }
        }
    }
}