A Bug's Life

                                             Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                 Total Submission(s): 13068    Accepted Submission(s): 4262

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     作為一個並查集的新手，加上也不是很努力，so，，兩天才解決了這個問題。。of course ，也是千辛萬苦看了網上的程式碼才馬馬虎虎“弄懂”這題。。。。。。。。。
相信也有很多像我一樣的新手不懂這題的思路。。那麼，就讓我來詳細地講一下這題吧。。
    首先，題意是：有一群‘bug’，它們跟人一樣，也分雌雄，也能交配，甚至也有同性戀。。。那麼，這題就是，通過給你一串資料，讓你判斷這群‘bug’中有沒有同性戀。
比如，輸入1 2，代表1和2進行了交配，那麼就是說明1和2是異性，這裡assume 1是雄，那麼2便是雌。緊接著輸入 2 3，代表2和3進行了交配，前面已經有了2是雌的，那麼3自然是雄的‘最後一對資料是 1 3，第一對中已經有1 是雄的，那麼3便是雌的咯，可是第二對資料中已經說明3是雄的，那麼與這對資料中3是雌的相矛盾。。so，唯一的解釋就是 3 是同性戀。。好啦，既然是同性戀，那麼便輸出 suspicious bugs found！
    思路：把每一個’bug‘看作一個節點，開兩個陣列，f[MAX]用來儲存每個節點的父節點，r[MAX]用來儲存每個節點的性別（這裡，性別用 1和0來表示，就好像人的男和女，動物的雌和雄一樣。開始把每一個節點的性別都設為0），那麼，如何來判斷 是 存在同性戀的呢？看下圖。（並查集的基本知識這裡就不重複強調了）


我的字確實很爛。。。不過自認為圖還是不錯滴
    那麼具體實現就直接看程式碼啦，，

#include<stdio.h>
int f[10001],r[10001],t,flag;

int find(int x)
{
    if(f[x]==x)
        return x;
    t=find(f[x]);
    r[x]=(r[f[x]]+r[x])&1;
    f[x]=t;
    return f[x];
}

void bin(int x,int y)
{
    int fx,fy;
    fx=find(x);
    fy=find(y);
    if(fx==fy)
    {
        if(r[x]==r[y])
            flag=1;
        return;
    }
    f[fx]=fy;
    r[fx]=(r[x]+r[y]+1)&1;
}

int main()
{
    int N,k=1;
    scanf("%d",&N);
    while(N--)
    {
        int n,s;
        int i;
        int x,y;
        flag=0;
        scanf("%d%d",&n,&s);
        for(i=0;i<=n;i++)
        {
            
            f[i]=i;
            r[i]=0;
        }
        for(i=1;i<=s;i++)
        {
            scanf("%d%d",&x,&y);
            if(flag==1)
                continue;
            bin(x,y);
        }
        printf("Scenario #%d:\n",k++);
        if(flag)printf("Suspicious bugs found!\n");
        else printf("No suspicious bugs found!\n");
        printf("\n");
    }
    return 0;
}