hdu 1829 并查集（食物链的弱化版）

http://acm.hdu.edu.cn/showproblem.php?pid=1829

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.

/**
hdu 1829  并查集（食物链的弱化版）
题目大意：给定n个昆虫，和m种恋爱关系，判断是否有同性恋
解题思路：并查集i-A表示i属于性别A若i-A和j-B,属于同一类，表示i为性别A和j为性别B是同时发生或者同时不发生。类似于poj1182
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=2005;
int par[maxn*2],_rank[maxn*2];

void init(int n)
{
    for(int i=0; i<=n; i++)
    {
        par[i]=i;
        _rank[i]=0;
    }
}

int find(int x)
{
    if(par[x]==x)
        return x;
    return par[x]=find(par[x]);
}

void unite(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)return;
    if(_rank[x]<_rank[y])
    {
        par[x]=y;
    }
    else
    {
        par[y]=x;
        if(_rank[x]==_rank[y])
            _rank[x]++;
    }
}

bool same(int x,int y)
{
    return find(x)==find(y);
}

int main()
{
    int T,tt=0,n,m;
    scanf("%d",&T);
    while(T--)
    {
        int flag=0;
        scanf("%d%d",&n,&m);
        init(n*2);
        while(m--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(same(x,y)||same(x+n,y+n))
            {
                flag=1;
            }
            else
            {
                unite(x,y+n);
                unite(x+n,y);
            }
        }
        printf("Scenario #%d:\n",++tt);
        if(flag)
            puts("Suspicious bugs found!\n");
        else
            puts("No suspicious bugs found!\n");
    }
    return 0;
}