POJ 2492 并查集应用的扩展

A Bug's Life

Time Limit: 10000MS Memory Limit: 65536K

Total Submissions: 28651 Accepted: 9331



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

/*********************************
     author   : Grant Yuan
     time     : 2014/10/5 11:29
     algorithm: 并查集应用的扩展
     source   : POJ 2492
**********************************/
#include<iostream>
#include<cstdio>
#define MAX 2007
using namespace std;
int parent[MAX],relation[MAX];
int n,t,m;
int find_parent(int k)
{
    if(k==parent[k])
        return k;
    int t=parent[k];
    parent[k]=find_parent(parent[k]);
    relation[k]=(relation[k]+relation[t]+1)%2;
    return parent[k];
}
void unite(int a,int b)
{
    int pa=find_parent(a);
    int pb=find_parent(b);
    parent[pa]=pb;
    relation[pa]=(relation[b]-relation[a])%2;
}
int main()
{
    scanf("%d",&t);
    int ct=1;bool first=0;
    while(t--){
        bool ans=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            parent[i]=i,relation[i]=1;
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            int pa=find_parent(a);
            int pb=find_parent(b);
            if(pa==pb){
                if(relation[a]==relation[b]) ans=1;
            }
            else unite(a,b);
        }
        printf("Scenario #%d:\n",ct++);
        if(ans) printf("Suspicious bugs found!\n\n");
        else printf("No suspicious bugs found!\n\n");
    }
    return 0;
}