A Bug‘s Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13709    Accepted Submission(s): 4449

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：给你若干个bug，保证他们是异性，看最后有没有同性恋‘

思路分析：基础并查集，维护与根节点的关系即可

poj1703 同类型

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=2000+100;
int fa[maxn];
int relate[maxn];//记录与父节点的性别关系，1代表异性，0代表同性
int n,m;
int kase=0;
int root(int x)
{
    if(x==fa[x]) return x;
    int t=root(fa[x]);
    relate[x]=(relate[x]+relate[fa[x]])%2;
    fa[x]=t;
    return fa[x];
}
void merge(int x,int y)
{
    int fx=root(x);
    int fy=root(y);
    if(fx==fy) return;
    fa[fx]=fy;
    relate[fx]=(relate[y]+relate[x]+1)%2;
    return;
}
int main()
{
    int T;
    int a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            fa[i]=i,relate[i]=0;
        bool flag=false;
        for(int j=1;j<=m;j++)
        {
            scanf("%d%d",&a,&b);
            if(root(a)==root(b)&&relate[a]==relate[b])
            {
                flag=true;
            }
            else merge(a,b);
            //printf("yes\n");
        }
        //printf("dsasd\n");
        printf("Scenario #%d:\n",++kase);
        if(flag) printf("Suspicious bugs found!\n\n");
        else printf("No suspicious bugs found!\n\n");

    }
    return 0;
}