hdu 1829 A Bug's Life (并查集)

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7176    Accepted Submission(s): 2330

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

 

Recommend

linle   |   We have carefully selected several similar problems for you:  1558 1198 1325 1213 2473 

 

//796MS    240K    1104 B    G++
/*

    题意：
        给出m对关系，每对都是异性，问有木有矛盾的
    
    并查集：
        开始自己写的集中思路都不对，其实就是要处理一个二分图，
    如果该二分图存在矛盾时就为有矛盾，具体思路：
        opp保存对立单位，set保并查集中的元素；
        当opp[x]==0 && opp[y]==0时，证明x、y无对立关系，设置对立关系：opp[x]=y;opp[y]=x;
        当opp[x]==0 && opp[y]!=0时，设置x对立关系，并合并opp[y]与x：opp[x]=y;merge(x,opp[y]);
        当opp[x]!=0 && opp[y]==0时，同上：opp[y]=x;merge(y,opp[x]); 
        当opp[x]!=0 && opp[y]!=0时，
            如find(x)==find(b)，证明存在矛盾
            否则将x、y与其对立关系的合并：merge(y,opp[x]);merge(x,opp[y]);  

*/
#include<stdio.h>
int set[2005];
int find(int x)
{
    if(x==set[x]) return x;
    return find(set[x]);    
}
void merge(int a,int b)
{
    set[find(a)]=find(b);
}
int main(void)
{
    int t,n,m;
    int x,y,k=1;
    scanf("%d",&t);
    while(t--)
    {
        int flag=0;
        int opp[2005]={0};
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) set[i]=i;
        for(int i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            if(!flag){
                if(opp[x]==0 && opp[y]==0){
                    opp[x]=y;opp[y]=x;
                }else if(opp[x]==0){
                    opp[x]=y;merge(x,opp[y]);
                }else if(opp[y]==0){
                    opp[y]=x;merge(y,opp[x]);
                }else{
                    if(find(x)==find(y)) flag=1;
                    merge(y,opp[x]);merge(x,opp[y]);
                }
            }
        }
        printf("Scenario #%d:
",k++);
        if(flag) printf("Suspicious bugs found!
");
        else printf("No suspicious bugs found!
");
        printf("
");
    }
    return 0;
}