A Bug's Life POJ - 2492 （并查集）

A Bug's Life

 POJ - 2492 

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!

题意：

判断给出的m对不同关系是否有矛盾

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
#define N 2010
#define nmax 6
#define inf 0x3f3f3f3f
int fa[N],col[N]; //col[x]表示x与根的关系，col[x]=0表示同性，col[x]=1表示异性
int fin(int x)
{
    if(fa[x]==x)
        return x;
    int t=fa[x];
    fa[x]=fin(fa[x]);
    col[x]=(col[x]+col[t])%2;
    return fa[x];
}
int judge(int x,int y)
{
    int fx=fin(x);
    int fy=fin(y);
    if(fx!=fy)
    {
        fa[fx]=fy;  //以fy为根,连接两个连通集合
        col[fx]=col[x]^col[y]^1;  
        /*把握一个初始条件，所有连通集合中，根节点r的col[r]==0；
        异或col[x]^col[y]的值若为0时，即x和y和根节点同性或都不同性，
        则若要保证x和y异性，就要把x,y的根节点关系置为异性，即
        col[x]^col[y]异或1就行；异或col[x]^col[y]的值若为0时同样分析，
        Ps：异或运算是多个if语句的精华
        */
    }
    else if(!(col[x]^col[y]))//与根节点的关系为相同，即x和y关系为同性
        return 0;
    return 1;
}
int main()
{
    int t,n,m,x,y,flag,k=1;
    scanf("%d",&t);
    while(t--)
    {
        flag=0;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            fa[i]=i;
            col[i]=0;
        }
        while(m--)
        {
            scanf("%d%d",&x,&y);
            if(flag)
                continue;
            if(!judge(x,y))
                flag=1;
        }
        printf("Scenario #%d:\n",k++);
        if(flag)
            printf("Suspicious bugs found!\n\n");
        else
            printf("No suspicious bugs found!\n\n");

    }
    return 0;
}