Lake Counting_深度搜索_递归

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30414		Accepted: 15195

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.

Source

USACO 2004 November

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感觉自己一直没有系统的训练，从网上买了一本挑战程序设计竞赛，开始挨着刷题，这是第一道。

#include <iostream>
#include <cstdio>

#define L 120

using namespace std;

int n,m;
char a[L][L];
int dx[]={,,,,,-,-,-};
int dy[]={,,,-,-,-,,};

void dfs(int x,int y){
    a[x][y]='.';
    for(int i=;i<;i++){

            int t=x+dx[i];
            int t2=y+dy[i];
            if(a[t][t2]=='W' && t>= && t<n && t2>= &&t2<m){
                dfs(t,t2);
            }
    }
}

int main()
{
    int cou=;
    while(~scanf("%d %d",&n,&m)){
        getchar();
        cou=;
        for(int i=;i<n;i++){
            for(int j=;j<m;j++){
                scanf("%c",&a[i][j]);
            }
            getchar();
        }
        for(int i=;i<n;i++){
            for(int j=;j<m;j++){
                if(a[i][j]=='W'){
                    dfs(i,j);
                    cou++;
                }
            }
        }
        printf("%d\n",cou);

    }
    return ;
}