*HDU1829 并查集
A Bug's Life
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14121 Accepted Submission(s): 4603
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.
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 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.
//将相同性别的bugs放进同一并查集里,这样只要输入的两个bug有相同的根就是gay.
//路径压缩的while 的find比递归的find稍快一些。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int t,n,m;
int fat[],sex[];
int find(int x)
{
int rt=x;
while(fat[rt]!=rt)
rt=fat[rt];
int i=x,j;
while(i!=rt)
{
j=fat[i];
fat[i]=rt;
i=j;
}
return rt;
}
/*int find(int x)
{
if(fat[x]!=x)
fat[x]=find(fat[x]);
return fat[x];
}*/
void connect(int x,int y)
{
int xx=find(x),yy=find(y);
if(xx!=yy)
fat[xx]=yy;
}
int main()
{
int a,b;
scanf("%d",&t);
for(int k=;k<=t;k++)
{
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
{
fat[i]=i;
sex[i]=;
}
bool flag=;
for(int i=;i<=m;i++)
{
scanf("%d%d",&a,&b);
if(flag) continue;
if(find(a)==find(b))
{
flag=;
continue;
}
if(sex[a]==) sex[a]=b;
else connect(sex[a],b);
if(sex[b]==) sex[b]=a;
else connect(sex[b],a);
}
printf("Scenario #%d:\n",k);
if(flag) printf("Suspicious bugs found!\n");
else printf("No suspicious bugs found!\n");
printf("\n");
}
return ;
}
*HDU1829 并查集的更多相关文章
- poj1182、hdu1829(并查集)
题目链接:http://poj.org/problem?id=1182 食物链 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: ...
- A Bug's Life(hdu1829种类并查集)
题意:有一群虫子,现在给你一些关系,判断这些关心有没有错 思路:向量种类并查集,下面讲一下向量的种类并查集 本题的各个集合的关心有两种0同性,1异性,怎么判断有错, 1.先判断他们是否在一个集合,即父 ...
- HDU1829(种类并查集)
ps:本来是想找个二分图判断的题来写,结果百度到这鬼题 Problem Description Background Professor Hopper is researching the sexua ...
- HDU1829【种类并查集】
题意: 检验给出条件是否有同性恋. 思路: 条件并查集. 还是一个类似的前缀和,sum[x]是x到根这段路径上的和,根一定是坐标越小的, 那么如果说对于同类(同一个集合)的判断就sum[a]是否等于s ...
- HDU-1829 A Bug's Life。并查集构造,与POJ1709异曲同工!
A Bug's Life Find them, Catch them 都是并查集构造的题,不久前 ...
- hdu1829 A Bug's Life(并查集)
开两个并查集.然后合并的时候要合并两次.这样在合并之前推断是否冲突,假设不冲突就进行合并,否则不须要继续合并. #include<cstdio> #include<cstdlib&g ...
- hdu 3038 How Many Answers Are Wrong(种类并查集)2009 Multi-University Training Contest 13
了解了种类并查集,同时还知道了一个小技巧,这道题就比较容易了. 其实这是我碰到的第一道种类并查集,实在不会,只好看着别人的代码写.最后半懂不懂的写完了.然后又和别人的代码进行比较,还是不懂,但还是交了 ...
- POJ 2492 A Bug's Life【并查集高级应用+类似食物链】
Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes ...
- BZOJ 4199: [Noi2015]品酒大会 [后缀数组 带权并查集]
4199: [Noi2015]品酒大会 UOJ:http://uoj.ac/problem/131 一年一度的“幻影阁夏日品酒大会”隆重开幕了.大会包含品尝和趣味挑战两个环节,分别向优胜者颁发“首席品 ...
随机推荐
- Servlet接口五种方法介绍
Servlet接口定义了5种方法: init() service() destroy() getServletConfig() getServletInfo() init() 在Servlet实例化后 ...
- Codeforces Round #262 (Div. 2) 1004
Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory lim ...
- PHP header函数使用大全
PHP header函数大全 header('Content-Type: text/html; charset=utf-8'); header('Location: http://52php.cnbl ...
- CMake命令/函数汇总(翻译自官方手册)
查看官方文档 cmake命令 选项 CMake变量 CMake命令汇总 / add_custom_command add_custom_target/add_definitions/add_depen ...
- Shell入门教程:流程控制(4)case 条件判断
case的语法结构: case 待测项 in 样式串1] 命令区域1 ;; (样式串2) 命令区域2 ;; 样式串3) 命令区域3 ;; *) 命令区域 ;; esac 命令区域,可以是单一指令或多行 ...
- JAVA设计模式--单例模式
单例设计模式 Singleton是一种创建型模式,指某个类采用Singleton模式,则在这个类被创建后,只可能产生一个实例供外部访问,并且提供一个全局的访问点. 核心知识点如下: (1) 将采用单例 ...
- Linux学习之一--VI编辑器的基本使用
vi编辑器是Linux系统下标准的编辑器.而且不逊色于其他任何最新的编辑器.可是会用的有多少呢.下面介绍一下vi编辑器的简单用法和部分命令.让你在Linux系统中畅行无阻. 基本上vi可以分为三种状态 ...
- 值得推荐的android开源框架
1.volley 项目地址https://github.com/smanikandan14/Volley-demo (1) JSON,图像等的异步下载: (2) 网络请求的排序(scheduling) ...
- Hibernate一对一、一对多、多对多注解映射配置
一对一: 一对多: 多对多:
- 我常用的find命令
查找某种类型文件中包含特定字符的文件 find /* -type f -name "*.php" |xargs grep "rename(" find ./|x ...