HDU-3081-Marriage Match 2(最大流, 二分答案, 并查集)
链接:
https://vjudge.net/problem/HDU-3081
题意:
Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids.
Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her boyfriend when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend.
Once every girl finds their boyfriends they will start a new round of this game—marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.
Now, here is the question for you, how many rounds can these 2n kids totally play this game?
思路:
二分枚举答案, 每次对图用枚举的答案建图,男女根据并查集确定能否配对连一条权值为1 的边.
跑最大流即可.
#代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 200+10;
const int INF = 1e9;
struct Edge
{
int from, to, cap;
};
vector<Edge> edges;
vector<int> G[MAXN];
int Dis[MAXN], Vis[MAXN];
int Fa[MAXN];
int Map[MAXN][MAXN];
int n, m, s, t, d;
void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge{from, to, cap});
edges.push_back(Edge{to, from, 0});
G[from].push_back(edges.size()-2);
G[to].push_back(edges.size()-1);
}
bool Bfs()
{
//Bfs构造分层网络
memset(Dis, -1, sizeof(Dis));
queue<int> que;
que.push(s);
Dis[s] = 0;
while (!que.empty())
{
int u = que.front();
que.pop();
// cout << u << endl;
for (int i = 0;i < G[u].size();i++)
{
Edge & e = edges[G[u][i]];
if (e.cap > 0 && Dis[e.to] == -1)
{
que.push(e.to);
Dis[e.to] = Dis[u]+1;
}
}
}
return (Dis[t] != -1);
}
int Dfs(int u, int flow)
{
//flow 表示当前流量上限
if (u == t)
return flow;
int res = 0;
for (int i = 0;i < G[u].size();i++)
{
Edge & e = edges[G[u][i]];
if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
{
int tmp = Dfs(e.to, min(flow, e.cap)); // 递归计算顶点 v
flow -= tmp;
e.cap -= tmp;
res += tmp;
edges[G[u][i]^1].cap += tmp;
if (flow == 0)
break;
}
}
if (res == 0)
Dis[u] = -1;
return res;
}
int MaxFlow()
{
int res = 0;
while (Bfs())
{
res += Dfs(s, INF);
}
return res;
}
int GetF(int x)
{
if (Fa[x] == x)
return x;
Fa[x] = GetF(Fa[x]);
return Fa[x];
}
bool Check(int mid)
{
for (int i = s;i <= t;i++)
G[i].clear();
edges.clear();
for (int i = 1;i <= n;i++)
{
AddEdge(s, i, mid);
AddEdge(n+i, t, mid);
}
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= n;j++)
{
if (Map[i][j])
AddEdge(i, n+j, 1);
}
}
int res = MaxFlow();
// cout << mid << ' ' << res << endl;
return res == mid*n;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int T;
cin >> T;
while (T--)
{
memset(Map, 0, sizeof(Map));
cin >> n >> m >> d;
for (int i = 1;i <= n;i++)
Fa[i] = i;
s = 0, t = n*2+1;
int u, v;
for (int i = 1;i <= m;i++)
{
cin >> u >> v;
Map[u][v] = 1;
}
for (int i = 1;i <= d;i++)
{
cin >> u >> v;
int tu = GetF(u);
int tv = GetF(v);
if (tu != tv)
Fa[tv] = tu;
}
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= n;j++)
{
if (i == j || GetF(i) != GetF(j))
continue;
for (int k = 1;k <= n;k++)
{
if (Map[i][k])
Map[j][k] = 1;
if (Map[j][k])
Map[i][k] = 1;
}
}
}
int l = 0, r = n;
int res = 0;
// Check(2);
while (l <= r)
{
int mid = (l+r)/2;
if (Check(mid))
{
res = max(res, mid);
l = mid+1;
}
else
r = mid-1;
}
cout << res << endl;
}
return 0;
}
HDU-3081-Marriage Match 2(最大流, 二分答案, 并查集)的更多相关文章
- HDU 3081 Marriage Match II 最大流OR二分匹配
Marriage Match IIHDU - 3081 题目大意:每个女孩子可以和没有与她或者是她的朋友有过争吵的男孩子交男朋友,现在玩一个游戏,每一轮每个女孩子都要交一个新的男朋友,问最多可以玩多少 ...
- HDU 3081 Marriage Match II (网络流,最大流,二分,并查集)
HDU 3081 Marriage Match II (网络流,最大流,二分,并查集) Description Presumably, you all have known the question ...
- HDU 3081 Marriage Match II(二分法+最大流量)
HDU 3081 Marriage Match II pid=3081" target="_blank" style="">题目链接 题意:n个 ...
- HDU 3081 Marriage Match II (二分图,并查集)
HDU 3081 Marriage Match II (二分图,并查集) Description Presumably, you all have known the question of stab ...
- [HNOI2006]公路修建问题 (二分答案,并查集)
题目链接 Solution 二分答案+并查集. 由于考虑到是要求花费的最小值,直接考虑到二分. 然后对于每一个二分出来的答案,模拟 \(Kruskal\) 的过程再做一遍连边. 同时用并查集维护联通块 ...
- HDU 3081 Marriage Match II 二分 + 网络流
Marriage Match II 题意:有n个男生,n个女生,现在有 f 条男生女生是朋友的关系, 现在有 m 条女生女生是朋友的关系, 朋友的朋友是朋友,现在进行 k 轮游戏,每轮游戏都要男生和女 ...
- HDU 3081 Marriage Match II (二分+网络流+并查集)
注意 这题需要注意的有几点. 首先板子要快,尽量使用带当前弧优化的dinic,这样跑起来不会超时. 使用弧优化的时候,如果源点设置成0,记得将cur数组从0开始更新,因为有的板子并不是. 其次这题是多 ...
- hdu 5652 India and China Origins(二分+bfs || 并查集)BestCoder Round #77 (div.2)
题意: 给一个n*m的矩阵作为地图,0为通路,1为阻碍.只能向上下左右四个方向走.每一年会在一个通路上长出一个阻碍,求第几年最上面一行与最下面一行会被隔开. 输入: 首行一个整数t,表示共有t组数据. ...
- 洛谷P1991无线通讯网[kruskal | 二分答案 并查集]
题目描述 国防部计划用无线网络连接若干个边防哨所.2 种不同的通讯技术用来搭建无线网络: 每个边防哨所都要配备无线电收发器:有一些哨所还可以增配卫星电话. 任意两个配备了一条卫星电话线路的哨所(两边都 ...
随机推荐
- vmware虚拟机设置时区、时间
首先查看时间发现和百度的时间不一样 [root@www ~]# dateWed Dec 5 14:00:32 CST 2018 1.配置ntp服务器,设置时区同步,请参照ntp篇 https://w ...
- Golang基础(7):go的net/rpc用法
一:PRC是什么? RPC(Remote Procedure Call) 远程过程调用,是一个计算通信协议.该协议允许一台计算机上的程序调用另外一台计算机上的程序.远程过程调用就是2个不在同一台计算机 ...
- cocos2dx基础篇(22) 基本动画CCAnimation/CCAnimate
[小知识] CCSpriteFrame :精灵帧. 它是相对动画而产生的,其实就是一张纹理图片. CCAnimationFrame :动画帧. 由精灵帧与间隔帧数组成,是动画CC ...
- 奥比中光Astra Pro在ROS系统中的使用
奥比中光Astra Pro在ROS系统中的使用 上一次介绍了Astra Pro在ubuntu中的使用,由于Astra Pro使用uvc传输彩色图像,所以当用rqt_image_view窗口 时,看不到 ...
- TIDB学习资料
TiDB 源码阅读系列文章(一)序 TiDB 源码阅读系列文章(二)初识 TiDB 源码 TiDB 源码阅读系列文章(三)SQL 的一生 TiDB 源码阅读系列文章(四)Insert 语句概览 TiD ...
- yii框架RBAC權限管理
基于角色的存取控制 (RBAC) 基于角色的存取控制 (RBAC) 提供了一个简单而强大的集中式存取控制机制. 详细的关于 RBAC 和诸多传统的存取控制方案对比的详情,请参阅 Wikipedia. ...
- array_chunk的用法和php操作大数据
一.array_chunk() 函数 二.php操作大数据 1.在操作大数量数据与数据库交互时,比如插入大量数据,db就会报错,这时可以把原本的数据用array_chunk分隔成几个数组块,再循环插入 ...
- __next__()
def f1(n): m=n while True: m+=1 yield m a=f1(5) print(a.__next__()) 结果:6
- centos 7下nginx搭建流媒体服务器【动态添加模块】
1.安装nginx依赖包 yum install gcc gcc-c++ openssl-devel zlib-devel pcre pcre-devel yamdi 2.下载解压nginx_mod_ ...
- 看电视剧<潜伏>有感
前几天看了老电视剧-潜伏,有一些感慨. 一,立场和真相都不重要,形式才是最重要的. 二.历史在不断的轮回中. 好汉历经千辛万苦杀掉了为害一方的恶霸,好汉的威望达到了顶峰,自然的成了村庄的守护者和掌控者 ...