A - Wireless Network POJ - 2236

题目大意:有n台坏掉的电脑,给出每台电脑的坐标,然后每次询问输入0(字符) x,表示电脑x恢复正常,输入S x y 询问x和y是否可以联网。只要是x和y的距离小于距离d,那么就可以联网,如果有个中介c使得x到c的距离小于d,y到c的距离小于d,那么x和y也可以联网。

题解:当修复好一台电脑后,然后判断与之前修好的电脑的距离,小于d的话,用并查集连在一起。(没敢这样想,感觉这样会T....)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=2E4+7;
int n,d;
int arrx[N],arry[N],pre[N];
int cnt[N];
int pos=0;
int dis(int a,int b){
    return pow(arrx[a]-arrx[b],2)+pow(arry[a]-arry[b],2)<=d*d;
}
int find(int x){
    return x==pre[x]? x:pre[x]=find(pre[x]);
}
void unite(int x,int y){
    x=find(x);y=find(y);
    pre[x]=y;
}
bool check(int x,int y){
    return find(x)==find(y);
}
int main(){
    cin>>n>>d;
    for(int i=1;i<=n;i++) cin>>arrx[i]>>arry[i];
    char s;
    for(int i=0;i<=n;i++) pre[i]=i;
    while(cin>>s){
        if(s==‘S‘){
            int x,y;
            scanf("%d%d",&x,&y);
            if(check(x,y)) cout<<"SUCCESS"<<endl;
            else cout<<"FAIL"<<endl;
        }
        else {
            int x;cin>>x;
            cnt[++pos]=x;
            for(int i=1;i<pos;i++){
                if(dis(x,cnt[i])){
                    unite(x,cnt[i]);
                }
            }
        }
    }
    return 0;
}

原文地址:https://www.cnblogs.com/Accepting/p/12639365.html

时间: 2024-11-13 06:50:28

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