并查集 例题：HDU 1829

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15115    Accepted Submission(s): 4917

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

23 31 22 31 34 21 23 4

 

Sample Output

Scenario #1:Suspicious bugs found!Scenario #2:No suspicious bugs found!
题意：就是两只相连的虫子（bug）一定是不同性别的，但是存在一种特殊的虫子是两性的（可以认为是雌雄同体，也可以认为是两性恋），给你T组数据，n只虫子，m个关系，问：有没有special bug？
思路：就是用并查集，如果原本两个独立的关系（集合），因为加 x-y 这条关系，可以将这两个集合连在一起，让其的pre[y]和pre[x]相同，也就是祖先相同（哈哈哈哈），这样就联系上有同一祖先的所有虫子的关系了。因为只有两性，所以，可以将其的深度对二取余判奇偶，同一祖先，还在同一深度（本是同一性别），它两还有关系，那就变态了哇！（special）
我的代码：
#include <iostream>#include<algorithm>#include<stdio.h>#include<cstring>using namespace std;#define maxx 100005int pre[maxx],rankk[maxx];int flag;int findp(int x){    if(x==pre[x])        return pre[x];    else    {        int t=findp(pre[x]);        rankk[x]=(rankk[pre[x]]+rankk[x])&1;        pre[x]=t;        return pre[x];    }}void mix(int u,int v){    int fu,fv;    fu=findp(u);    fv=findp(v);    if(fu==fv)    {        if(rankk[u]==rankk[v])            flag=1;        return;    }    pre[fu]=fv;    rankk[fu]=(rankk[u]+rankk[v]+1)&1;}int main(){    int n,m;    int T;    int t=1;    cin>>T;    while(T--)    {        cin>>n>>m;        int u,v;        flag=0;        for(int i=1; i<=n; i++)        {            pre[i]=i;            rankk[i]=0;        }        for(int i=0; i<m; i++)        {            cin>>u>>v;            mix(u,v);        }        printf("Scenario #%d:\n",t++);        if(flag==1)            printf("Suspicious bugs found!\n");        else            printf("No suspicious bugs found!\n");            cout<<endl;    }    return 0;}



这里有一片详细的并查集介绍：http://www.cnblogs.com/cyjb/p/UnionFindSets.html