HDU 1829 A Bug's Life

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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.

并查集的一道题目;

题目大概就是2种性别 ,不同性别的可以交配,相同的性别如果交配则出现特殊性别;

第一步: 初始化是基本的步骤;

第二步: 先枚举几个例子,来找出子节点 与其父节点的父节点 的关系

第三步:将给出的2个节点 分好性别关系

主函数主要是加以判断来输出结果




#include <stdio.h>//存储的是其父亲的下表int bugs[2010];int relation[2010];//1:相同性别 0:不同性别//初始化void init(int len){    for(int i = 0;i <= len; i++)    {        bugs[i] = i;        relation[i] = 1;    }}//找到根int find(int bug){    if(bugs[bug]==bug)return bug;    int tem = bugs[bug];    bugs[bug] = find(bugs[bug]);//递归更新域,返回最终的父亲节点,把所有的孩子都更新了    //注意这里,求当前位置和父亲的关系,记录之前父亲的位置为tem,然后因为是递归,    //此时的relation[tem]已经在递归中更新过了,也就是孩子和父亲的关系+父亲和爷爷的关系+1然后模2就得到    //孩子和爷爷的关系,这里用0和1表示,0表示不同性别,1表示相同性别    relation[bug] = (relation[bug]+relation[tem]+1)%2;    return bugs[bug];}void union_set(int a,int b,int x,int y){    //合并,让前边的集合的根指向后边集合的根,成为一个集合    bugs[x]=y;    //更新前边集合根和新的集合根之间的关系,    //注意这里,relation[a]+relation[x]与relation[b]    //相对于新的父节点必须相差1个等级,因为他们不是gay    relation[x] = (relation[b]-relation[a])%2;}int main(){    int S;    int n,inter;    int bug1,bug2,parent1,parent2;    bool flag;//false:无同性恋,true:有同性恋    scanf("%d",&S);    for(int i=1; i<=S;i++)    {        scanf("%d%d",&n,&inter);        flag = false;        init(n);//初始化,使其父节点为自己        for(int j = 1; j <= inter; j++)        {            scanf("%d%d",&bug1,&bug2);            if(flag)continue;            parent1 = find(bug1);            parent2 = find(bug2);            if(parent1==parent2)            {                if(relation[bug1]==relation[bug2])//同性                flag = true;            }            union_set(bug1,bug2,parent1,parent2);        }             if(flag)        printf("Scenario #%d:\nSuspicious bugs found!\n",i);        else        printf("Scenario #%d:\nNo suspicious bugs found!\n",i);        printf("\n");    }    return 0;}



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.
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