POJ-2492-A Bug's Life（并查集分类）

最新推荐文章于 2019-09-12 21:47:04 发布

转载最新推荐文章于 2019-09-12 21:47:04 发布 · 124 阅读

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原文链接：http://www.cnblogs.com/yzhhh/p/9943191.html

本文探讨了一种利用并查集算法解决虫族互动谜题的方法，通过二倍节点技巧实现集合分类，确保实验中不存在同性虫之间的互动，验证了关于虫族性别的假设。

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!

思路：把性别不同的的放在不同集合里，如果出现已经存在，并且和一起输进来的节点同性，则可以断定出现异常虫。

技巧：二倍节点，从而实现集合分类。

坑点：这个题要压缩并查集，将所有节点直接指向编号节点，否则会TE。

#include<cstdio>
using namespace std;
int n,m;
int bin[4010];

//这样写find会超时 
//int find(int x){
//    while(bin[x]!=x)
//        x=bin[x];
//    return x;    
//}

int find(int x) 
{
    if(bin[x] == x)
        return bin[x];
    return bin[x] = find(bin[x]);
}
void merge(int x,int y){
    int fx,fy;
    fx=find(x);
    fy=find(y);
    if(bin[fx]!=bin[fy])
        bin[fx]=fy;
}

bool same(int x,int y){
    return find(x)==find(y);
}

int main(){
    int t ;
    scanf("%d",&t);
    for(int p=1;p<=t;p++){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n*2;i++)
            bin[i]=i;
        int flag=0;        
        while(m--){
            int x,y;
            scanf("%d%d",&x,&y);
                        
            if(same(x,y)||same(x+n,y+n)) flag=1;
            if(flag) break;    
            merge(x,y+n);
            merge(x+n,y);
        }
        printf("Scenario #%d:\n",p);
        if(flag)
            printf("Suspicious bugs found!\n\n");
        else 
            printf("No suspicious bugs found!\n\n");
        
    }
    return 0;
}