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): 14610    Accepted Submission(s): 4771

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.

题目的大概意思就是找输入的几组数据中是否矛盾，如果一组数据中1喜欢2，2喜欢3，而1又喜欢3，则矛盾

用一个数组存下虫子之间的关系，然后再判断是否有重复的关系就可以了。1为同性，0为异性

AC代码来自博客，因为他写的很好很全，就套用他的了

http://www.cnblogs.com/newpanderking/archive/2012/10/12/2721799.html

#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:
Suspicious bugs found!
",i);
        else
        printf("Scenario #%d:
No suspicious bugs found!
",i);
        printf("
");
    }
    return 0;
}