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题目链接：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!
题目大意：Hopper教授正在研究一种稀有虫子的交配行为。他假设它们有两种性别并且它们只与异性交配。在他的实验中，很容易识别虫子和它们的交配行为，因为虫子背后印有编号。
问题
给定一组虫子的交配行为，确定实验是支持教授的假设即虫子没有同性恋，还是有部分交配行为不符合假设。
这道题实际上还是并查集，和以往不同的是。以往给出两个元素的关系，然后认为这两个元素有联系，把它们划归为同一个集合。 最后看看能划分出多少个不同的集合。 本道题，可以认为有两个集合，异性和同性。 给出关系，希望你检查这两个集合有没有连接，即有没有同性恋。
大体思路： ‘合并’操作， ‘查找’操作不变。设置一个sex[] 数组，sex[i]存储与 i 性别相反的对象编号 ,初始化为0 。 然后将于i 发生联系的元素合并为同一个集合（因为人为它们是同性）。下次输入x,y时，如果 find(x)==fidn(y)，我们就可以人为出现了反常行为。
总之，就是将同性的化为同一个集合中去。
代码如下：
#include <cstdio>
#include <cstring>
#define N 2017
int sex[N], set[N];
int find(int x)
{
return x == set[x]?x:set[x]=find(set[x]);
}
void Union(int x, int y)
{
int f1 = find(x);
int f2 = find(y);
if(f1 != f2)
{
set[f2] = f1;
}
}
void init(int n)
{
int i;
for(i = 1; i <= n; i++)
{
set[i] = i;
sex[i] = 0;
}
}
int main()
{
int n, m;
int t, i, j;
int x, y, cas = 0, flag;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init(n);
flag = 0;
for(i = 0; i < m; i++)
{
scanf("%d%d",&x,&y);
if(flag)
continue;
if(find(x) == find(y))
{
flag = 1;
continue;
}
if(sex[x] == 0)
sex[x] = y;
else
Union(sex[x],y);
if(sex[y] == 0)
sex[y] = x;
else
Union(sex[y],x);
}
printf("Scenario #%d:\n",++cas);
if(flag)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
printf("\n");
}
return 0;
}
hdu1829 A Bug's Life （并查集）