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
- 如何在Ubuntu / CentOS 6.x上安装Bugzilla 4.4
这里,我们将展示如何在一台Ubuntu 14.04或CentOS 6.5/7上安装Bugzilla.Bugzilla是一款基于web,用来记录跟踪缺陷数据库的bug跟踪软件,它同时是一款免费及开源软件 ...
- oracle分页排序,点击下一页数据不刷新
oracle数据库中,如果每一页的最后一条和次页第一条数据的排序字段重复,会导致排序混乱,出现点击下一页数据不刷新的现象,所以一般排序至少选择一个相对唯一的字段.在前端页面可以输入排序条件的场景中,最 ...
- win10序列号 2019年10月测试
win10序列号 N3415-266GF-AH13H-WA3UE-5HBT4 win10序列号 NPK3G-4Q81M-X4A61-D553L-NV68D win10序列号 N617H-84K11-6 ...
- mysql标准规范
一.基础规范 表存储引擎必须使用InnoDB 表字符集默认使用utf8,必要时候使用utf8mb4 解读: (1)通用,无乱码风险,汉字3字节,英文1字节 (2)utf8mb4是utf8的超集,有存储 ...
- JS 截取字符的方法
substr() 方法 substr() 方法可在字符串中抽取从 start 下标开始的指定数目的字符. stringObject.substr(start,length) start:必需.要抽取的 ...
- for循环练习题:拆解字符并输入下标
test = input('请输入:') for item in range(0,len(test)): print(item,test[item])
- Leetcode 38.报数 By Python
报数序列是一个整数序列,按照其中的整数的顺序进行报数,得到下一个数.其前五项如下: 1. 1 2. 11 3. 21 4. 1211 5. 111221 1 被读作 "one 1" ...
- CentOS7 nginx 最简单的安装以及设置开机启动
1. 下载tar包. 2. 解压缩tar包 3. 安装必须的部分 yum包 yum install -y gcc pcre pcre-devel openssl openssl-devel gd gd ...
- 洛谷 P3368 树状数组 题解
题面 本题随便看两眼就知道该题满足了优美的查分性质: 对于在区间[x,y]内操作时,应该将查分数组的第x项和第y+1项进行相反操作: 询问答案时,问第i个数的值就是查分数组的前i项和: 暴力+玄学卡常 ...
- pgsql物理复制(pgsql 备库的搭建以及角色互换,提升)
结构图如下: Postgresql早在9.0版本开始支持物理复制,也称为流复制,通过从实例级复制出一个与主库一模一样的备库.流复制同步方式有同步,异步两种,如果主节点和备节点不是很忙,通常异步模式下备 ...