POJ 2492 A Bug's Life （种类并查集）

A Bug's Life

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 52818		Accepted: 16983

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!

题目大意与分析

n个虫子，m个关系，每个关系代表a和b能交配，要求找出有没有同性恋虫子

这个题和POJ1703类似，也是用种类并查集做，对于输入的ab，查询是否属于同一个祖先即可

 

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <math.h>
#include <cstdio>
using namespace std;

int s[4005],t,n,m,k=0,flag;

int findf(int x)
{
    return x==s[x]?x:s[x]=findf(s[x]);
}

void hebing(int x,int y)
{
    int fx=findf(x);
    int fy=findf(y);
    if(fx!=fy)
    {
        s[fx]=fy;
    }
}

int main()
{
    cin>>t;
    while(t--)
    {
        k++;
        flag=0;
        cin>>n>>m;
        for(int i=1;i<=2*n;i++)
        {
            s[i]=i;
        }
        while(m--)
        {
            int a,b;
            cin>>a>>b;
            if(findf(a)==findf(b))
            {
                flag=1;
            }
            else
            {
                hebing(a,b+n);
                hebing(a+n,b);
            }
        }
        cout<<"Scenario #"<<k<<":"<<endl;
        if(flag==1)
        {
            cout<<"Suspicious bugs found!"<<endl;
        }
        else
        {
            cout<<"No suspicious bugs found!"<<endl;
        }
        cout<<endl;
    }
}