hdu 1829 A Bug's Life（分组并查集（偏移量））

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9204    Accepted Submission(s): 2961

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

题意：

     1喜欢2,2喜欢3，而1又喜欢3，则矛盾。 （排出同性恋）

基础并查集的新应用：

      分组并查集（偏移量）：开一个并查集，但是要加个偏移向量数组，来记录每个节点距离根节点的距离

    代码：

/*带权值的并查集*/
#include<cstdio>
#define maxn 2005
int father[maxn],rank[maxn];
int tt,nn,mm;
void  init()
{
   for(int i=1;i<nn;i++){
       father[i]=i;
       rank[i]=0;
  }
}

int fin(int x)
{
  if(x==father[x])   return x;
  int temp=fin(father[x]);
  rank[x]=(rank[x]+rank[father[x]])%2;
  father[x]=temp;
  return father[x];
}
int unin(int a,int b)
{
    int x=fin(a);
    int y=fin(b);
    if(x==y)
    {
        if(rank[a]==rank[b])return 1;  //说明矛盾
        return 0;
    }
    father[x]=y;
    rank[x]=(rank[a]+rank[b]+1)%2;
    return 0;
}
int main()
{
   int a,b;
   int tag;
  scanf("%d",&tt);
  for(int j=1;j<=tt;j++)
  {
      tag=0;
    scanf("%d%d",&nn,&mm);
    init();
    for(int i=0;i<mm;i++)
    {
       scanf("%d%d",&a,&b);
       tag+=unin(a,b);
    }
    printf("Scenario #%d:
",j);

    if(tag)puts("Suspicious bugs found!");
    else puts("No suspicious bugs found!");
    puts("");
  }
  return 0;
}