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;
}