HDU-1829.A Bug's Life(带权并查集)

博客围绕A Bug’s Life问题展开,Hopper教授研究稀有虫子性行为,假设分两性且只与异性互动。给出虫子互动列表,需判断实验是否支持该假设。介绍了输入输出格式,还提及用带权并查集解决此问题。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A Bug’s Life

Problem Description

Background

Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.

Problem

Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

Output

The output for every scenario is a line containing “Scenario #i:”, where i is the number of the scenario starting at 1, followed by one line saying either “No suspicious bugs found!” if the experiment is consistent with his assumption about the bugs’ sexual behavior, or “Suspicious bugs found!” if Professor Hopper’s assumption is definitely wrong.

Sample Input

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

带权并查集

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <unordered_set>
#include <set>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <set>
#include <queue>
#include <cmath>
using namespace std;
const int maxn = 1000010;
int fa[maxn];
int ranks[maxn]; //0表示同类,1表示异类
int m,n; //m只虫子,n种行为
int findroot(int x)
{
    if(x!=fa[x])
    {
        int root=findroot(fa[x]);
        ranks[x]=(ranks[x]+ranks[fa[x]])%2;
        fa[x]=root;
    }
    return fa[x];
}
int main()
{
    int t,ans=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        bool flag = true;
        memset(ranks,0,sizeof(ranks));
        for(int i=0;i<=m;i++)
            fa[i]=i;
        int u,v;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            if(flag==false) continue; //如果已经找到可疑,不必再检查后面的
            int du=findroot(u);
            int dv=findroot(v);
            if(du==dv)
            {
                if((ranks[u]-ranks[v])%2==0)
                    flag=false;
            }
            else
            {
                fa[du]=dv;
                ranks[du]=(ranks[u]-ranks[v]+1)%2;
            }
        }
        if(!flag)
        {
            printf("Scenario #%d:\n",ans++);
            printf("Suspicious bugs found!\n");
        }
        else
        {
            printf("Scenario #%d:\n", ans++);
            printf("No suspicious bugs found!\n");
        }
        printf("\n");
    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值