A Bug‘s Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10063    Accepted Submission(s): 3288

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.

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意：

t组数据，每组n个人，m对有暧昧关系的，问是否存在 gay ？

代码如下：

//考察知识点：分组并查集，听说有两种办法，但是只会这一种
//相较于一般的并查集，分组并查集，在查找父节点的函数方面没有做改动，
//在合并的时候 出现了选择性。主要改变在于merge函数上面。
//变动为
/*
void merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y-max);
	if(fx==fy)
	{
		flag=0;
		return ;
	}
	fy=find(y);
	if(fx!=fy)
	father[fx]=fy;
}
*/ 

#include<stdio.h>
#define max 2020
int father[2*max];
int flag;
int find(int x)
{
	int r=x;
	while(r!=father[r])
	r=father[r];
	return r;
}
void merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y-max);
	if(fx==fy)
	{
		flag=0;
		return ;
	}
	fy=find(y);
	if(fx!=fy)
	{
		father[fx]=fy;
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	int a,b,m,n,count=1;
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=max+n;++i)
		{
			father[i]=i;
		}
		flag=1;
		for(int i=1;i<=m;++i)
		{
			scanf("%d%d",&a,&b);
			if(flag)
			{
				merge(a,b+max);
				merge(b,a+max);
			}
		}
		printf("Scenario #%d:\n",count++);
		if(flag)
		{
			printf("No suspicious bugs found!\n\n");
		}
		else
		printf("Suspicious bugs found!\n\n");
	}
	return 0;
}