ACMA并查集 Bug's Life

最新推荐文章于 2020-10-19 15:39:33 发布

原创最新推荐文章于 2020-10-19 15:39:33 发布 · 110 阅读

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本文介绍了一种通过记录虫群互动来验证其性别行为假设的方法。利用并查集算法将虫子分为两类，并检查是否存在不符合预期的同性互动。

题目：

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.

题意：emmmm给出哪两个可以发生性行为，判断是否有同性恋

分析：食物链简单版本，，只有两个分类，用2*n数组储存，分成两堆，之后判断a很a+n是否是在一块；

代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int f[6005];
int t,n,m;
void intt()
{
for(int i=1;i<=n*2;i++)
f[i]=i;
}
int getf(int v)
{
if(f[v]==v) return v;
else
{
f[v]=getf(f[v]);
return f[v];
}
}
bool same(int a,int b)
{
return getf(a)==getf(b);
}
void unit(int a,int b)
{
int t1,t2;
t1=getf(a);
t2=getf(b);
if(t1!=t2)
f[t2]=t1;
return ;
}
int main()
{
int i,a,b,ans,casee=0;
scanf("%d",&t);
while(t--)
{
ans=0;
scanf("%d %d",&n,&m);
intt();
for(i=1;i<=m;i++)
{
scanf("%d %d",&a,&b);
unit(a,b+n);
unit(b,a+n);
if(same(a,a+n)||same(b,b+n))
ans=1;
}

if(ans!=0)
printf("Scenario #%d: \nSuspicious bugs found!\n",++casee);
else
printf("Scenario #%d: \nNo suspicious bugs found!\n",++casee);
printf("\n");
}
return 0;
}