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
————————————————————————————————
题目的意思是给出n个士兵和几组关系,士兵两两配对搭档,问最后有多少人有搭档并
输出
思路:一般图匹配模板题,套带花树开花算法模板
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std; #define LL long long
const int INF = 0x3f3f3f3f; const int MAXN = 250;
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 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 = 1; tu <= N; tu++)
if(InBlossom[Base[tu]])
{
Base[tu] = NewBase;
if(!InQueue[tu]) Push(tu);
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,0,sizeof(Father));
for(int i = 1; i <= N; i++)
Base[i] = i;
Head = Tail = 1;
Push(Start);
Finish = 0;
while(Head < Tail)
{
int u = Pop();
for(int v = 1; v <= N; v++)
if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v))
{
if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0))
BloosomContract(u,v);
else if(Father[v] == 0)
{
Father[v] = u;
if(Match[v] > 0)
Push(Match[v]);
else
{
Finish = v;
return;
}
}
}
}
}
void AugmentPath()
{
int u,v,w;
u = Finish;
while(u > 0)
{
v = Father[u];
w = Match[v];
Match[v] = u;
Match[u] = v;
u = w;
}
}
void Edmonds()
{
memset(Match,0,sizeof(Match));
for(int u = 1; u <= N; u++)
if(Match[u] == 0)
{
Start = u;
FindAugmentingPath();
if(Finish > 0)AugmentPath();
}
} int main()
{ int u,v;
while(~scanf("%d",&N))
{
memset(Graph,false,sizeof(Graph)); while(~scanf("%d%d",&u,&v))
{
Graph[u][v] = Graph[v][u] = true;
} Edmonds();//进行匹配
int cnt=0;
for(int i=1; i<=N; i++)
if(Match[i]>0)
cnt++;
printf("%d\n",cnt);
for(int i=1; i<=N; i++)
if(i<Match[i])
printf("%d %d\n",i,Match[i]); } return 0;
}

  

URAL1099. Work Scheduling(一般图匹配带花树开花算法)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. URAL 1099. Work Scheduling (一般图匹配带花树)

    1099. Work Scheduling Time limit: 0.5 secondMemory limit: 64 MB There is certain amount of night gua ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 【UOJ 79】 一般图最大匹配 (✿带花树开花)

    从前一个和谐的班级,所有人都是搞OI的.有 n 个是男生,有 0 个是女生.男生编号分别为 1,…,n. 现在老师想把他们分成若干个两人小组写动态仙人掌,一个人负责搬砖另一个人负责吐槽.每个人至多属于 ...

  7. 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 ...

  8. 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 ...

  9. 【learning】一般图最大匹配——带花树

    问题描述 ​ 对于一个图\(G(V,E)\),当点对集\(S\)满足任意\((u,v)\in S\),均有\(u,v\in V,(u,v)\in E\),且\(S\)中没有点重复出现,我们称\(S\) ...

随机推荐

  1. 20.Mysql锁机制

    20.锁问题锁是计算机协调多个进程或线程并发访问某一资源的机制. 20.1 Mysql锁概述锁类型分为表级锁.页面锁.行级锁.表级锁:一个线程对表进行DML时会锁住整张表,其它线程只能读该表,如果要写 ...

  2. opencv中imread第二个参数的意义

    文档中是这么写的: Flags specifying the color type of a loaded image: CV_LOAD_IMAGE_ANYDEPTH - If set, return ...

  3. BZOJ1057或洛谷1169 [ZJOI2007]棋盘制作

    BZOJ原题链接 洛谷原题链接 设\(L[i][j],R[i][j],H[i][j]\)表示点\((i,j)\)向左.右.上尽量拓展的左端点.右端点.上端点的坐标. \(L,R\)直接初始化好,\(H ...

  4. Luogu 2812 校园网络 - Tarjan

    Description 给出一个有向图, 要求出至少从哪几个点出发, 能不漏地经过所有节点. 再求出至少加几条边, 才能使图变成一个强联通分量 Solution 求出所有强联通分量, 形成一个有向无环 ...

  5. soapui groovy脚本汇总

    出处:https://www.jianshu.com/p/ce6f8a1f66f4 一.一些内部元件的访问 testRunner.testCase开头 1.向下访问 testRunner.testCa ...

  6. Python编程笔记(第二篇)二进制、字符编码、数据类型

    一.二进制 bin() 在python中可以用bin()内置函数获取一个十进制的数的二进制 计算机容量单位 8bit = 1 bytes 字节,最小的存储单位,1bytes缩写为1B 1KB = 10 ...

  7. GBDT原理

    样本编号 花萼长度(cm) 花萼宽度(cm) 花瓣长度(cm) 花瓣宽度 花的种类 1 5.1 3.5 1.4 0.2 山鸢尾 2 4.9 3.0 1.4 0.2 山鸢尾 3 7.0 3.2 4.7 ...

  8. Arithmetic Slices II - Subsequence LT446

    446. Arithmetic Slices II - Subsequence Hard A sequence of numbers is called arithmetic if it consis ...

  9. pc-H5 适配方案

    一.介绍 在前端项目页面开发中,尤其是H5页面开发,我们常常要适配各种分辨率的屏幕,才能让用户获得最好的体验效果.pc也是如此,很多页面是一屏,也是要适配各种尺寸的分辨率.这时候我们就需要对各种分辨率 ...

  10. mybatis 为什么要设置jdbcType

    转载自:http://makemyownlife.iteye.com/blog/1610021 前天遇到一个问题 异常显示如下: 引用 Exception in thread "main&q ...