带权并查集 hdu1829

最新推荐文章于 2023-08-03 20:43:59 发布

苯上的甲基

最新推荐文章于 2023-08-03 20:43:59 发布

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分类专栏：并查集文章标签：并查集

本文链接：https://blog.youkuaiyun.com/qq_40620465/article/details/83928663

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本文介绍了一种使用带权并查集解决的算法问题，源自教授霍珀对一种稀有昆虫性行为的研究。实验中，昆虫被编号以便观察其交互行为，目标是验证昆虫是否只与异性互动。通过构建带权并查集，算法能够判断实验数据是否支持教授的假设，即昆虫社会中不存在同性恋现象。文章详细解释了权重转移方程，并提供了完整的C++代码实现。

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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

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

Hint

Huge input,scanf is recommended.

其实本题就是个带权并查集，根节点相同且权值相同，那么肯定就是同性。

关键是权值的转移方程，权值的转移可以用向量表示，详情可以见我之前的博客。(https://blog.youkuaiyun.com/qq_40620465/article/details/83342386)

设雄性为权值=1 雌性=0

p[x]->y=p[x]->x+x->y+y->p[y]

r[y]=r[x]+1+r[y] 因为x,y之间能交配说明肯定有一个是雄性那么x,y的权值和肯定为1

r[]数组保存的是父节点与子节点的关系权值

#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <ctype.h>
#define  LL long long
#define  ULL unsigned long long
#define mod 1000000007
#define INF 0x7ffffff
#define mem(a,b) memset(a,b,sizeof(a))
#define MODD(a,b) (((a%b)+b)%b)
using namespace std;
const int N=1000005;
int p[N],r[N];
bool flag;
int findfa(int x)
{
    if(x!=p[x]){
        int fa=findfa(p[x]);
        r[x]=(r[p[x]]+r[x])&1;//父节点之间的权值转移
        p[x]=fa;
    }
    return p[x];
}
void Union(int x,int y)
{
    int fax=findfa(x),fay=findfa(y);
    if(fax==fay){
        if(r[x]==r[y]) flag=true;
        return ;
    }
    p[fay]=fax;
    r[fay]=(r[x]+1+r[y])&1;

}
int main()
{
   int T;
   int cot=1;
   scanf("%d",&T);
   while(T--){
    int n,m;
    flag=false;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) p[i]=i,r[i]=0;

    for(int i=0;i<m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        if(flag) continue;
        Union(x,y);

    }
    printf("Scenario #%d:\n",cot++);
    if(flag) printf("Suspicious bugs found!\n");
    else printf("No suspicious bugs found!\n");
    printf("\n");
   }
   

    return 0;


}