HDU 2444 The Accomodation of Students(判断二分图+最大匹配)
The Accomodation of Students
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3775 Accepted Submission(s): 1771
Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.
Calculate the maximum number of pairs that can be arranged into these double rooms.
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.
Proceed to the end of file.
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6
3
/* ***********************************************
Author :pk28
Created Time :2015/10/6 19:53:21
File Name :hdu2444.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 40000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std; bool cmp(int a,int b){
return a>b;
}
int n;
int m,l,tot;
int pre[],match[];
int vis[maxn]; struct node{
int v,next;
}edge[maxn];
void init(){
l=;
memset(pre,-,sizeof pre);
memset(match,-,sizeof match);
}
void add(int u,int v){
edge[l].v=v;
edge[l].next=pre[u];
pre[u]=l++;
}
int dfs(int u){
for(int i=pre[u];i+;i=edge[i].next){
int v=edge[i].v;
if(!vis[v]){
vis[v]=;
if(match[v]==-||dfs(match[v])){
match[v]=u;
return ;
}
}
}
return ;
}
int hungary(){
tot=;
for(int i=;i<=n;i++){
cle(vis);
if(dfs(i))tot++;
}
return tot;
}
int color[maxn];
bool bs(int u){
for(int i=pre[u];i+;i=edge[i].next){
int v=edge[i].v;
if(color[v]==color[u])return false;
if(!color[v]){
color[v]=-color[u];
if(!bs(v))return false;
}
}
return true;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int a,b;
while(cin>>n>>m){
init(); for(int i=;i<=m;i++){
scanf("%d%d",&a,&b);
add(a,b);
add(b,a);
}
cle(color);
color[]=;
if(!bs())puts("No");
else printf("%d\n",hungary()/);
}
return ;
}
HDU 2444 The Accomodation of Students(判断二分图+最大匹配)的更多相关文章
- hdu 2444 The Accomodation of Students 判断二分图+二分匹配
The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K ( ...
- hdu 2444 The Accomodation of Students (判断二分图,最大匹配)
The Accomodation of StudentsTime Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (J ...
- HDU 2444 The Accomodation of Students【二分图最大匹配问题】
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2444 题意:首先判断所有的人可不可以分成互不认识的两部分.如果可以分成 ,则求两部分最多相互认识的对数. ...
- (hdu)2444 The Accomodation of Students 判断二分图+最大匹配数
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2444 Problem Description There are a group of s ...
- HDU 2444 The Accomodation of Students (二分图最大匹配+二分图染色)
[题目链接]:pid=2444">click here~~ [题目大意]: 给出N个人和M对关系,表示a和b认识,把N个人分成两组,同组间随意俩人互不认识.若不能分成两组输出No,否则 ...
- hdu 2444 The Accomodation of Students 判断是否构成二分图 + 最大匹配
此题就是求最大匹配.不过需要判断是否构成二分图.判断的方法是人选一点标记为红色(0),与它相邻的点标记为黑色(1),产生矛盾就无法构成二分图.声明一个vis[],初始化为-1.通过深搜,相邻的点不满足 ...
- HDU 2444 The Accomodation of Students(二分图判定+最大匹配)
这是一个基础的二分图,题意比较好理解,给出n个人,其中有m对互不了解的人,先让我们判断能不能把这n对分成两部分,这就用到的二分图的判断方法了,二分图是没有由奇数条边构成环的图,这里用bfs染色法就可以 ...
- hdu 2444 The Accomodation of Students 【二分图匹配】
There are a group of students. Some of them may know each other, while others don't. For example, A ...
- HDU 2444 The Accomodation of Students 二分图判定+最大匹配
题目来源:HDU 2444 The Accomodation of Students 题意:n个人能否够分成2组 每组的人不能相互认识 就是二分图判定 能够分成2组 每组选一个2个人认识能够去一个双人 ...
随机推荐
- net8:文本文件的创建及其读写
原文发布时间为:2008-08-06 -- 来源于本人的百度文章 [由搬家工具导入] using System;using System.Data;using System.Configuration ...
- LeetCode OJ--Implement strStr()
http://oj.leetcode.com/problems/implement-strstr/ 判断一个串是否为另一个串的子串 比较简单的方法,复杂度为O(m*n),另外还可以用KMP时间复杂度为 ...
- HUNAN 11569 Just Another Knapsack Problem(AC自动机+dp)
http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11569&courseid=0 给出目标串,每个子串和 ...
- HTTP请求的缓存(Cache)机制
原文地址:http://small.aiweimeng.top/index.php/archives/58.html 先来一张图: ####下面简单的来描述一下HTTP Cache机制: 当资源资源第 ...
- T1013 求先序排列 codevs
http://codevs.cn/problem/1013/ 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 黄金 Gold 题解 查看运行结果 题目描述 Descr ...
- luogu P1886 滑动窗口(单调队列
题目描述 现在有一堆数字共N个数字(N<=10^6),以及一个大小为k的窗口.现在这个从左边开始向右滑动,每次滑动一个单位,求出每次滑动后窗口中的最大值和最小值. 例如: The array i ...
- Hadoop安装和基本单机部署
下载安装 # 下载 $ cd /usr/local $ wget http://mirrors.hust.edu.cn/apache/hadoop/common/hadoop-2.9.2/hadoo ...
- 转:C#并口热敏小票打印机打印位图
最近一直在研究并口小票打印机打印图片问题,这也是第一次和硬件打交道,不过还好,最终成功了. 这是DEMO的窗体: 下面是打印所需要调用的代码: class LptControl { private s ...
- hdoj-1856-More is better【并查集】
More is better Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others) To ...
- 转:NetBeans的远程Linux C开发实践
转: http://blog.csdn.net/jacktan/article/details/9268535 一直以来总觉得NetBeans生活在Eclipse的阴影下,同样做为一款不错的基于Jav ...