hdu A Bug's Life

 

A Bug's Life

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

Total Submission(s): 353    Accepted Submission(s): 133

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

分析：路径压缩的并查集。用给点相对于根节点的距离%2表示是同性还是异性。merge()中修改被吞并树根的值使之满足所给两点的差距为奇数，find()中利用路径压缩将该树相对于已改变的根节点的奇偶值修改正确。

 

#include<cstdio>
#include<cstring>
int fa[2010],rank[2010];
int n,flag;
int find(int x)
{
    if(fa[x]==x)
        return x;
    int t;
    t=find(fa[x]);
    rank[x]=(rank[x]+rank[fa[x]])%2;
    fa[x]=t;
    return t;
}
void merge(int x,int y)
{
    int fx,fy;
    fx=find(x);
    fy=find(y);
  //  printf("x=%d y=%d fx=%d fy=%d\n",x,y,fx,fy);
    if(fx==fy)
    {
        if(rank[x]==rank[y])
        {
            flag=0;
        }
    }
    else
    {
        fa[fx]=fy;
        rank[fx]=(rank[x]+rank[y]+1)%2;
    }
}
int main()
{
    int T,m,i,a,b,t;
    scanf("%d",&T);
    for(t=1;t<=T;++t)
    {
        scanf("%d%d",&n,&m);
        flag=1;
        memset(rank,0,sizeof(rank));
        for(i=1;i<=n;++i)
        {
            fa[i]=i;
        }
        while(m--)
        {
            scanf("%d%d",&a,&b);
            if(!flag)
                continue;
            merge(a,b);
        }
        printf("Scenario #%d:\n",t);
        if(flag)
            printf("No suspicious bugs found!\n\n");
        else
            printf("Suspicious bugs found!\n\n");
    }
    return 0;
}