1099. Work Scheduling

Time limit: 0.5 second
Memory limit: 64 MB
There is certain amount of night guards that are available to protect the local junkyard from possible junk robberies. These guards need to scheduled in pairs, so that each pair guards at different night. The junkyard CEO ordered you to write a program which given the guards characteristics determines the maximum amount of scheduled guards (the rest will be fired). Please note that each guard can be scheduled with only one of his colleagues and no guard can work alone.

Input

The first line of the input contains one number N ≤ 222 which is the amount of night guards. Unlimited number of lines consisting of unordered pairs (ij) follow, each such pair means that guard #i and guard #j can work together, because it is possible to find uniforms that suit both of them (The junkyard uses different parts of uniforms for different guards i.e. helmets, pants, jackets. It is impossible to put small helmet on a guard with a big head or big shoes on guard with small feet). The input ends with Eof.

Output

You should output one possible optimal assignment. On the first line of the output write the even number C, the amount of scheduled guards. Then output C/2 lines, each containing 2 integers (ij) that denote that i and j will work together.

Sample

input output
3
1 2
2 3
1 3
2
1 2

模板题!!!

以下是模板:

 /* ***********************************************
Author :kuangbin
Created Time :2013/8/21 22:56:05
File Name :F:\2013ACM练习\专题学习\图论\一般图匹配带花树\URAL1099.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std; const int MAXN = ;
int N; //点的个数,点的编号从1到N
bool Graph[MAXN][MAXN];
int Match[MAXN];
bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN];
int Head,Tail;
int Queue[MAXN];
int Start,Finish;
int NewBase;
int Father[MAXN],Base[MAXN];
int Count;//匹配数,匹配对数是Count/2
void CreateGraph()
{
int u,v;
memset(Graph,false,sizeof(Graph));
scanf("%d",&N);
while(scanf("%d%d",&u,&v) == )
{
Graph[u][v] = Graph[v][u] = true;
}
}
void Push(int u)
{
Queue[Tail] = u;
Tail++;
InQueue[u] = true;
}
int Pop()
{
int res = Queue[Head];
Head++;
return res;
}
int FindCommonAncestor(int u,int v)
{
memset(InPath,false,sizeof(InPath));
while(true)
{
u = Base[u];
InPath[u] = true;
if(u == Start) break;
u = Father[Match[u]];
}
while(true)
{
v = Base[v];
if(InPath[v])break;
v = Father[Match[v]];
}
return v;
}
void ResetTrace(int u)
{
int v;
while(Base[u] != NewBase)
{
v = Match[u];
InBlossom[Base[u]] = InBlossom[Base[v]] = true;
u = Father[v];
if(Base[u] != NewBase) Father[u] = v;
}
}
void BloosomContract(int u,int v)
{
NewBase = FindCommonAncestor(u,v);
memset(InBlossom,false,sizeof(InBlossom));
ResetTrace(u);
ResetTrace(v);
if(Base[u] != NewBase) Father[u] = v;
if(Base[v] != NewBase) Father[v] = u;
for(int tu = ; tu <= N; tu++)
if(InBlossom[Base[tu]])
{
Base[tu] = NewBase;
if(!InQueue[tu]) Push(tu);
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,,sizeof(Father));
for(int i = ;i <= N;i++)
Base[i] = i;
Head = Tail = ;
Push(Start);
Finish = ;
while(Head < Tail)
{
int u = Pop();
for(int v = ; v <= N; v++)
if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v))
{
if((v == Start) || ((Match[v] > ) && Father[Match[v]] > ))
BloosomContract(u,v);
else if(Father[v] == )
{
Father[v] = u;
if(Match[v] > )
Push(Match[v]);
else
{
Finish = v;
return;
}
}
}
}
}
void AugmentPath()
{
int u,v,w;
u = Finish;
while(u > )
{
v = Father[u];
w = Match[v];
Match[v] = u;
Match[u] = v;
u = w;
}
}
void Edmonds()
{
memset(Match,,sizeof(Match));
for(int u = ; u <= N; u++)
if(Match[u] == )
{
Start = u;
FindAugmentingPath();
if(Finish > )AugmentPath();
}
}
void PrintMatch()
{
Count = ;
for(int u = ; u <= N;u++)
if(Match[u] > )
Count++;
printf("%d\n",Count);
for(int u = ; u <= N; u++)
if(u < Match[u])
printf("%d %d\n",u,Match[u]);
}
int main()
{
CreateGraph();//建图
Edmonds();//进行匹配
PrintMatch();//输出匹配数和匹配
return ;
}

URAL 1099. Work Scheduling (一般图匹配带花树)的更多相关文章

  1. URAL1099 Work Scheduling —— 一般图匹配带花树

    题目链接:https://vjudge.net/problem/URAL-1099 1099. Work Scheduling Time limit: 0.5 secondMemory limit: ...

  2. kuangbin带你飞 匹配问题 二分匹配 + 二分图多重匹配 + 二分图最大权匹配 + 一般图匹配带花树

    二分匹配:二分图的一些性质 二分图又称作二部图,是图论中的一种特殊模型. 设G=(V,E)是一个无向图,如果顶点V可分割为两个互不相交的子集(A,B),并且图中的每条边(i,j)所关联的两个顶点i和j ...

  3. HDU 4687 Boke and Tsukkomi (一般图匹配带花树)

    Boke and Tsukkomi Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Othe ...

  4. URAL 1099 Work scheduling 一般图的最大匹配 带花树算法(模板)

    R - Work scheduling Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u ...

  5. URAL 1099 Work Scheduling (一般图最大匹配) 模板题【带花树】

    <题目链接> <转载于 >>>  > 题目大意: 给出n个士兵,再给出多组士兵之间两两可以匹配的关系.已知某个士兵最多只能与一个士兵匹配.求最多能够有多少对匹 ...

  6. URAL1099. Work Scheduling(一般图匹配带花树开花算法)

    1099. Work Scheduling Time limit: 0.5 second Memory limit: 64 MB There is certain amount of night gu ...

  7. HDU 4687 Boke and Tsukkomi 一般图匹配,带花树,思路,输出注意空行 难度:4

    http://acm.hdu.edu.cn/showproblem.php?pid=4687 此题求哪些边在任何一般图极大匹配中都无用,对于任意一条边i,设i的两个端点分别为si,ti, 则任意一个极 ...

  8. HDOJ 4687 Boke and Tsukkomi 一般图最大匹配带花树+暴力

    一般图最大匹配带花树+暴力: 先算最大匹配 C1 在枚举每一条边,去掉和这条边两个端点有关的边.....再跑Edmonds得到匹配C2 假设C2+2==C1则这条边再某个最大匹配中 Boke and ...

  9. ZOJ 3316 Game 一般图最大匹配带花树

    一般图最大匹配带花树: 建图后,计算最大匹配数. 假设有一个联通块不是完美匹配,先手就能够走那个没被匹配到的点.后手不论怎么走,都必定走到一个被匹配的点上.先手就能够顺着这个交错路走下去,最后一定是后 ...

随机推荐

  1. java.lang.IllegalArgumentException: Page directive: invalid value for import

    我的项目原来用的tomcat版本是apache-tomcat-7.0.53,后来为了安全原因将版本升至\apache-tomcat-7.0.57,发现有的jsp页面出现下面的异常: java.lang ...

  2. 最全Pycharm教程(26)——Pycharm搜索导航之文件名、符号名搜索(转)

    1.准备一个工程 向你的工程中添加一个Python文件,并输入一些源码,例如: 2.转到对应文件.类.符号 Pycharm提供的一个很强力的功能就是能够根据名称跳转到任何文件.类.符号所在定义位置. ...

  3. POJ 1392 Ouroboros Snake(数位欧拉)

    题目链接:http://poj.org/problem?id=1392 题目大意:题意看的我头痛,其实跟HDU2894差不多,但是这题要求输出这条路径上第k个数,而不是输出路径. 解题思路:也跟HDU ...

  4. Hive入门学习随笔(一)

    Hive入门学习随笔(一) ===什么是Hive? 它可以来保存我们的数据,Hive的数据仓库与传统意义上的数据仓库还有区别. Hive跟传统方式是不一样的,Hive是建立在Hadoop HDFS基础 ...

  5. DB2和Oracle中Date比较

  6. 洛谷P2692 覆盖 题解

    题目传送门 这道题一开始想使用二维的bool型数组来存,最后统计.但看到数据范围... 所以就改用两个bool型数组(一维),分别储存横.列,最后将横.列面积求出来,再减去重复算的面积(横的个数*列的 ...

  7. Python爬虫笔记(一)

    个人笔记,仅适合个人使用(大部分摘抄自python修行路) 1.爬虫Response的内容 便是所要获取的页面内容,类型可能是HTML,Json(json数据处理链接)字符串,二进制数据(图片或者视频 ...

  8. CodeForces 805E Ice cream coloring

    直觉,构造. 画了几个样例,发现可以随便构造......先构造根节点的完全图,每个点置为不同的颜色,然后构造儿子节点的完全图...... #include <cstdio> #includ ...

  9. 实现RMQ的两种常用方法

    RMQ RMQ(Range Maximum/Minimum Question)是指区间最值问题,在OI中较为常见,一般可以用ST表和线段树实现. ST表是基于倍增思想的一种打表方法,在确定区间范围和所 ...

  10. JAVA基础关键字小结一

    基础部分总是看起来简单,若要明白原理和使用场景,需要慢慢的体会. 本文对常用的关键字如final,transient,volatile,static以及foreach循环的原理重新梳理了一遍. 一.f ...