HDU_1829_ABug'sLife

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11844    Accepted Submission(s): 3859

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

23 31 22 31 34 21 23 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

 

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linle

貌似这个问题属于带权并查集

但是我就生生的使用了集合的关系

首先所有昆虫应该属于两个集合

但是仅从每组输入，我们并不知道昆虫属于哪个集合

因此只能给他们各自一个集合

如果A B然后又有A C我们可以判断B与C属于一个集合

如果我们发现两个属于同一个集合，或者两个的对立集合是同一个集合

则两者同性，发生bug

不发生bug的两个发生关系，那么应该合并a和b的对立集合，b和a的对立集合

具体代码见下

#include <iostream>#include <stdio.h>#include <map>using namespace std;const int M=2005;map<int,int> uf;        //昆虫所在集合map<int,int> op;        //昆虫对立性别所在集合int find(int i){    if(i==uf[i])        return i;    return uf[i]=find(uf[i]);}void merge(int i,int j){    int t1=find(i);    int t2=find(j);    if(t1==t2)        return;    uf[t2]=t1;    merge(op[t2],op[t1]);   //同时要合并对立集合}int main(){    //freopen("1.in","r",stdin);    int n,m;    int t;    int co=1;    int i1,i2;    int f;    scanf("%d",&t);    while(t--)    {        f=0;        uf.clear();        op.clear();        scanf("%d%d",&n,&m);        for(int i=0;i<m;i++)        {            scanf("%d%d",&i1,&i2);            if(f)                continue;            if(!uf[i1])                uf[i1]=i1;            if(!uf[i2])                uf[i2]=i2;            if(find(i1)==find(i2))                f=1;            else            {                int par1=find(i1);                int par2=find(i2);                if(!op[par1])                    op[par1]=par2;                if(!op[par2])                    op[par2]=par1;                if(find(op[par1])==find(op[par2]))                    f=1;                else                {                    merge(op[par1],i2);                    merge(op[par2],i1);                }            }        }        printf("Scenario #%d:\n",co++);        if(f)            printf("Suspicious bugs found!\n\n");        else            printf("No suspicious bugs found!\n\n");    }    return 0;}