洛谷 P1596 [USACO10OCT]湖计数Lake Counting

题目链接

https://www.luogu.org/problemnew/show/P1596


题目描述

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.

由于近期的降雨,雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连,而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图,确定当中有多少水坑。


输入输出格式

输入格式:

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.

第1行:两个空格隔开的整数:N 和 M 第2行到第N+1行:每行M个字符,每个字符是'W'或'.',它们表示网格图中的一排。字符之间没有空格。

输出格式: 

Line 1: The number of ponds in Farmer John's field.

一行:水坑的数量


思路

没错,我又找到了一个可以暴搜的题目,因为又是判断连通块

似乎并没有什么技术含量的暴搜,有时候,就可以A掉一个题!!

8个if,让你的人生更加美好!!

因为所有的水坑都是连在一起的,所以我们可以for循环找到某一个水坑的某一个元素,将ans加1,然后将其周围的水全部变成陆地(及将这个水坑删除),然后这样一步一步搜下来就可以找出水坑数啦!

接下来我们就看代码吧!


代码

#include<iostream>
#include<cstdio>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<deque>
using namespace std;

int n,m;
char a[101][101];
bool f[101][101]= {false};

int baosou(int x,int y) {//将访问的这个水坑删除
    if(a[x][y+1]=='W') {
        a[x][y+1]='.';
        baosou(x,y+1);
    }
    if(a[x][y-1]=='W'){
        a[x][y-1]='.';
        baosou(x,y-1);
    }
    if(a[x+1][y]=='W') {
        a[x+1][y]='.';
        baosou(x+1,y);
    }
    if(a[x-1][y]=='W') {
        a[x-1][y]='.';
        baosou(x-1,y);
    }
    if(a[x+1][y+1]=='W') {
        a[x+1][y+1]='.';
        baosou(x+1,y+1);
    }
    if(a[x+1][y-1]=='W') {
        a[x+1][y-1]='.';
        baosou(x+1,y-1);
    }
    if(a[x-1][y+1]=='W') {
        a[x-1][y+1]='.';
        baosou(x-1,y+1);
    }
    if(a[x-1][y-1]=='W') {
        a[x-1][y-1]='.';
        baosou(x-1,y-1);
    }
}

int main() {
    scanf("%d%d",&n,&m);
    if(m==0)return 0;
    int ans=0;
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            cin>>a[i][j];
            //听我的,永远不要用scanf输入字符
        }
    }
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            if(a[i][j]=='W') {
                a[i][j]='.';
                ans++;//找到之后将这个点变成旱地并将ans加1
                baosou(i,j);//开始暴搜
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

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转载自www.cnblogs.com/loceaner/p/10732465.html