链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1829
原题：
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.
分析与总结：
做了这题才再次发现自己对并查集的了解真的只是一些皮毛，并查集的很多高级应用我都还不懂，而这题只是种类并查集中最简单的 = =|||。
假设有
1，2
2，3
即1--2--3， 明显相邻的两个不能是同性别的，如果相邻两个是同性的，那么说明就可能是有同性恋存在。
本栏目更多精彩内容：http://www.bianceng.cn/Programming/sjjg/
上面假设1是男性，用false来代替，那么按顺序分别是false, true, false
假设 再加个关系3， 1
那么由于1和3已经都有值了，都是false，说明可能有同性恋。
种类并查集的关键在于与结点与根结点的距离， 如果距离是奇数那么性别就和跟结点相反，如果是偶数就和跟结点性别相同。
代码：
#include<cstdio>
#define N 2005
using namespace std;
int f[N],rank[N], n, k;
bool flag;
inline void init(){
flag=false;
for(int i=0; i<=n; ++i)
f[i]=i, rank[i]=0;
}
int find(int x){
if(x==f[x])return f[x];
int t=find(f[x]);
rank[x] = (rank[f[x]]+rank[x])&1;
f[x]=t;
return f[x];
}
void Union(int x, int y){
int a=find(x), b=find(y);
if(a==b){
if(rank[x]==rank[y])
flag=true;
return;
}
f[a]=b;
rank[a] = (rank[x]+rank[y]+1)&1;
}
int main(){
int T,a,b,cas=1;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&k);
init();
for(int i=0; i<k; ++i){
scanf("%d%d",&a,&b);
if(flag)continue;
Union(a,b);
}
printf("Scenario #%d:\n",cas++);
if(flag)printf("Suspicious bugs found!\n");
else printf("No suspicious bugs found!\n");
printf("\n");
}
return 0;
}
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