学步园

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.

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.

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.

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

Huge input,scanf is recommended.

TUD Programming Contest 2005, Darmstadt, Germany

重新做了一下这题，以前用的是染色法判断是否有某条边的两端颜色一样

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

const int N=2000010;
const int maxn=2010;
int col[maxn];

struct node
{
	int next;
	int to;
}edge[N];
int head[maxn];
int tot;

void addedge(int from,int to)
{
	edge[tot].to=to;
	edge[tot].next=head[from];
	head[from]=tot++;
}

bool bfs(int root)
{
	queue<int>qu;
	while(!qu.empty())
		qu.pop();
	col[root]=1;
	qu.push(root);
	while(!qu.empty())
	{
		int u=qu.front();
		qu.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].to;
			if(col[v]==-1)
			{
				col[v]=(!col[u]);
				qu.push(v);
			}
			if(col[v]==col[u])
				return false; 
		}
	}
	return true;
}

int main()
{
	int t;
	scanf("%d",&t);
	int n,m,u,v;
	int icase=1;
	while(t--)
	{
		memset(head,-1,sizeof(head));
		memset(col,-1,sizeof(col));
		tot=0;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&u,&v);
			addedge(u,v);
			addedge(v,u);
		}
		bool flag;
		for(int i=1;i<=n;i++)
		{
			if(col[i]==-1)
			{
				flag=bfs(i);
				if(!flag)
					break;
			}
		}
		printf("Scenario #%d:\n",icase++);
		if(flag)
			printf("No suspicious bugs found!\n\n");
		else
			printf("Suspicious bugs found!\n\n");
	}
	return 0;
}

现在用了一下并查集
设opp[x] = y表示x的对面是y，某个x可能有多个对立面，我们合并这些对立面，那么显然这些对立面是同性的，因此如果输入的2个值是在同一个集合里，就说明它们是同性恋，否则就根据情况来合并集合

/*************************************************************************
    > File Name: POJ2492.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2015年01月22日 星期四 13时19分31秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 2010;
int opp[N];
int father[N];

int find (int x)
{
	if (father[x] == -1)
	{
		return x;
	}
	return father[x] = find (father[x]);
}

void merge (int a, int b)
{
	int aa = find (a);
	int bb = find (b);
	if (aa != bb)
	{
		father[aa] = bb;
	}
}

int main ()
{
	int t;
	int icase = 1;
	scanf("%d", &t);
	int n, m;
	while (t--)
	{
		int u, v;
		bool flag = false;
		scanf("%d%d", &n, &m);
		memset (father, -1, sizeof(father));
		memset (opp ,-1, sizeof(opp));
		while (m--)
		{
			scanf("%d%d", &u, &v);
			if (flag)
			{
				continue;
			}
			int uu = find (u);
			int vv = find (v);
			if (uu == vv)
			{
				flag = true;
			}
			else
			{
				if (opp[u] == - 1 && opp[v] == -1)
				{
					opp[u] = v;
					opp[v] = u;
				}
				else if (opp[u] == -1)
				{
					opp[u] = v;
					merge (u, opp[v]);
				}
				else if (opp[v] == -1)
				{
					opp[v] = u;
					merge (v, opp[u]);
				}
				else
				{
					merge (v, opp[u]);
					merge (u, opp[v]);
				}
			}
		}
		printf("Scenario #%d:\n", icase++);
		if (flag)
		{
			printf("Suspicious bugs found!\n");
		}
		else
		{
			printf("No suspicious bugs found!\n");
		}
		printf("\n");
	}
	return 0;
}