码迷,mamicode.com
首页 > 其他好文 > 详细

POJ2492 A Bug's Life (easy)

时间:2014-11-05 16:53:41      阅读:137      评论:0      收藏:0      [点我收藏+]

标签:des   io   ar   os   for   sp   strong   div   on   

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. 


题目大意:n个昆虫,m组关系,每组关系给出的两个昆虫属于不同的性别,判断是否有Suspicious(连poj都这么。。。美丽。。。?)

思路:类似与团伙的题目,简单并查集的应用。。。

code:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int fa[2001]={0},enemy[2001]={0};
int rool(int x)
{
if (fa[x]!=x) fa[x]=rool(fa[x]);
return fa[x];
}
int main()
{
int i,j,t,n,m,ci,a,b,r1,r2;
bool f;
cin>>t;
for (ci=1;ci<=t;++ci)
{
scanf("%d%d",&n,&m);
memset(enemy,0,sizeof(enemy));
for (i=1;i<=n;++i)
 fa[i]=i;
f=false;
for (i=1;i<=m;++i)
{
scanf("%d%d",&a,&b);
if (rool(a)==rool(b)) f=true;
if (!f)
{
 if (enemy[a]==0) enemy[a]=b;
 else 
 {
r1=rool(enemy[a]);
r2=rool(b);
fa[r1]=r2;
 }
 if (enemy[b]==0) enemy[b]=a;
 else
 {
r1=rool(a);
r2=rool(enemy[b]);
fa[r1]=r2;
 }
   }
}
printf("%s%d%s\n","Scenario #",ci,":");
if (!f) printf("%s\n\n","No suspicious bugs found!");
else printf("%s\n\n","Suspicious bugs found!");
}

 

POJ2492 A Bug's Life (easy)

标签:des   io   ar   os   for   sp   strong   div   on   

原文地址:http://www.cnblogs.com/Rivendell/p/4076359.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!