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HDU 1829 A Bug's Life 并查集

2016-10-06 21:34 387 查看

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1025 Accepted Submission(s): 376

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

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

HintHuge input,scanf is recommended.

Source

题目大意：

给出 n 个人 m 个信息，然后给出 a b ，a b 性别不同。然后问：有没有矛盾。

思路：

相当于建立两个并查集，同性放在一起。

如果 a 为 boy 那么 a+n 为 girl 。然后就这样了。

AC代码：

#include <iostream>
#include<stdio.h>
using namespace std;
const int MAX = 30000;
int n,m,k;
int parent[MAX+10];
int GetParent(int a)
{
if( parent[a]!= a)
parent[a] = GetParent(parent[a]);
return parent[a];
}
void Merge(int a,int b)
{
int p1 = GetParent(a);
int p2 = GetParent(b);
if( p1 == p2 )
return;
parent[p2] = p1;
}
int main()
{
int T;
scanf("%d",&T);
int kk=1;
while(T--)
{
cout<<"Scenario #"<<kk<<":"<<endl;
scanf("%d%d",&n,&m);
for(int i = 0;i <= 2*n; ++i)
{
parent[i] = i;
}
int work=0;
for(int i = 0;i < m; ++i)
{
int h,s;
scanf("%d",&h);
scanf("%d",&s);
if(work)
continue;
if(GetParent(h)==GetParent(s))
work=1;
else
{
Merge(h,s+n);
Merge(s,h+n);
}
}
if(work)
cout<<"Suspicious bugs found!"<<endl<<endl;
else
cout<<"No suspicious bugs found!"<<endl<<endl;
kk++;
}
return 0;
}

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标签： hdu 并查集

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