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. (转)Android学习路线总结,绝对干货

    一.前言 不知不觉自己已经做了几年开发了,由记得刚出来工作的时候感觉自己能牛逼,现在回想起来感觉好无知.懂的越多的时候你才会发现懂的越少. 如果你的知识是一个圆,当你的圆越大时,圆外面的世界也就越大. ...

  2. spec文件写作规范

    spec文件写作规范 2008-09-28 11:52:17 分类: LINUX 1.The RPM system assumes five RPM directories BUILD:rpmbuil ...

  3. 洛谷4054 [JSOI2009]计数问题

    原题链接 二维树状数组模板题. 对每一种颜色开一棵二维树状数组统计即可. #include<cstdio> using namespace std; const int N = 310; ...

  4. Eclipse新建tld文件

    tld(tag lib description文件)就是以.tld结尾的XML文件 选好目录右键 --> New --> Other -->找到XML FIle --> Nex ...

  5. [Robot Framework] 调用ExcelLibrary

    安装ExcelLibrary for Robot Framework 参考 : http://navinet.github.io/robotframework-excellibrary/ 打开wind ...

  6. coocsCreator杂记

    判断是否继承 cc.isChildClassOf = function (subclass, superclass) { 获取所有super classes CCClass.getInheritanc ...

  7. ubuntu系统ftp连接 以及ssh连接

    tfp连接 ssh连接 ubuntu下ssh使用 与 SCP 使用 1 ssh远程登录服务器 ssh username@remote_ip #将username换成自己的用户名,将remote_ip换 ...

  8. Python 单列

    1.__new__内置方法 在对类进行实例化时自动执行 功能1:为对象分配空间 功能2:返回空间的引用 2.单列实现方法 class MusicPlayer: # 记录对象内存引用,初始值为None ...

  9. 第五周 PSP 燃尽图 以及 进度条总结

    1.PSP DATE START-TIME END-TIME EVENT           DELTA TYPE 3.12 9.30 11.30 环境搭建 音乐30min QQ25min       ...

  10. mybatis学习 十六 auto_mapping实现连表查询

    只能使用多表联合查询方式. 要求:查询出的列别和属性名相同. 点字符  "."  在 SQL 是关键字符,两侧添加反单引号(Tab键上的一个字符) <select id=&q ...