Lake Counting_深度搜索_递归

Posted 2020-08-14 多一份不为什么的坚持

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[8]={0,1,1,1,0,-1,-1,-1};
int dy[8]={1,1,0,-1,-1,-1,0,1};

void dfs(int x,int y){
    a[x][y]=‘.‘;
    for(int i=0;i<8;i++){

            int t=x+dx[i];
            int t2=y+dy[i];
            if(a[t][t2]==‘W‘ && t>=0 && t<n && t2>=0 &&t2<m){
                dfs(t,t2);
            }
    }
}

int main()
{
    int cou=0;
    while(~scanf("%d %d",&n,&m)){
        getchar();
        cou=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                scanf("%c",&a[i][j]);
            }
            getchar();
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(a[i][j]==‘W‘){
                    dfs(i,j);
                    cou++;
                }
            }
        }
        printf("%d\n",cou);


    }
    return 0;
}