A Bug's Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19429 Accepted Submission(s): 6206

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!

Hint

Huge input,scanf is recommended.

Source

TUD Programming Contest 2005, Darmstadt, Germany

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题目意思：

Bug有两种性别，异性之间才交往，让你根据数据判断是否存在同性恋，输入有 t 组数据,每组数据给出bug数量n, 和关系数m, 以下m行给出相交往的一对Bug编号 a, b。只需要判断有没有，按题目要求输出

分析：

两个种类（两性）

种类并查集

判断存在同性恋的标准：

根节点相同的且性别相同的则是同性恋。

code：

#include <iostream>
#include<algorithm>
#include <cstdio>
#include<cstring>
#include<math.h>
#include<memory>
using namespace std;
typedef long long LL;
#define max_v 2005
int pa[max_v];//前驱
int re[max_v];//性别
int flag;
void make_set(int x)
{
    pa[x]=x;
    re[x]=0;
}
int find_set(int x)
{
    if(x==pa[x])
        return pa[x];
    int t=find_set(pa[x]);
    re[x]=(re[pa[x]]+re[x])%2;
    pa[x]=t;
    return pa[x];
}
void union_set(int x,int y)
{
    int a=find_set(x);
    int b=find_set(y);
    if(a==b)
    {
        if(re[x]==re[y])
            flag=1;//根节点相同的且性别相同的则是同性恋。
        return ;
    }
    pa[a]=b;
    re[a]=(re[x]-re[y]+3)%2;//保证 0 1
}
int main()
{
    int t,j=1;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        flag=0;
        for(int i=1;i<=n;i++)
            make_set(i);
        for(int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            if(flag==1)
                continue;
            union_set(a,b);
        }
        printf("Scenario #%d:
",j++);
        if(flag==1)
             printf("Suspicious bugs found!
");
        else
           printf("No suspicious bugs found!
");
        printf("
");
    }
    return 0;
}