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HDU 1829 A Bug’s Life(分组并查集)

本文由 Ocrosoft 于 2016-07-24 22:43:36 发表

A Bug’s Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13717    Accepted Submission(s): 4453

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.
 

Solution

连着两题并查集高级应用,题解都看不懂了…

题意:给出a b这样的数据,表示a和b相爱了,问里面有没有同性恋。

举例:
1 2
2 3
3 1
假设1 2异性,2 3异性,那么1 3同性,但是第三组数据是3 1,数据冲突,可以认为1 3是同性恋(又不用找出谁是同性恋,有就行了)。

方法一:开两个并查集,这种没写。可以百度HDU 1829第一个题解就是。

方法二:用一个relation数组记录(爷孙)性别关系,在find里更新relation。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 2000+10
using namespace std;
int n,m,pre[MAXN],relation[MAXN];
int find(int x)
{
    //if(pre[x]!=x)pre[x]=find(pre[x]);
    //return pre[x];
    if(pre[x]==x)return x;
    int p=pre[x];
    pre[x]=find(pre[x]);
    relation[x]=(relation[x]+relation[p]+1)%2;
    return pre[x];
}
void Union(int x,int y)
{
    int fx=find(x),fy=find(y);
    if(fx!=fy)pre[fx]=fy;
    relation[fx]=(relation[y]-relation[x])%2;
}
void init(int n=MAXN-1)
{
    for(int i=0; i<=n; i++)pre[i]=i,relation[i]=1;
}
int main()
{
    int N;
    cin>>N;
    for(int I=1; I<=N; I++)
    {
        bool flag=0;
        cin>>n>>m;
        init(n);
        for(int i=0; i<m; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(flag)continue;
            int fx=find(x),fy=find(y);
            if(fx==fy&&relation[x]==relation[y])flag=1;
            Union(x,y);
        }
        if(flag)printf("Scenario #%d:\nSuspicious bugs found!\n\n",I);
        else printf("Scenario #%d:\nNo suspicious bugs found!\n\n",I);
    }
    return 0;
}

 

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