Marriage Match II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5147 Accepted Submission(s): 1665
Problem Description
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?
Input
There are several test cases. First is a integer T, means the number of test cases.
Each test case starts with three integer n, m and f in a line (3<=n<=100,0<m<n*n,0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.
Output
For each case, output a number in one line. The maximal number of Marriage Match the children can play.
Sample Input
1
4 5 2
1 1
2 3
3 2
4 2
4 4
1 4
2 3
Sample Output
2
Author
starvae
Source
HDU 2nd “Vegetable-Birds Cup” Programming Open Contest
最大流 + 二分
二分最大匹配数
并查集处理朋友关系
设最大匹配数为k
建图时，s = 0,t = 2*n+1
s 向所有 女孩（1-n) 连一条容量为k的边
所有男孩向 t (n+1- 2*n) 连一条容量为k的边
女孩向能与之配对的男孩连一条容量为1的边
然后跑最大流 判断是否 满流 即 f = n*k
如果满流则代表可配对方式至少有k种，然后二分------
p.s 这道题也可以用二分图匹配做，以后补
然后虽然这样建图我勉强能接受了，但是还是有点不理解，因为能有k种一定能满流n*k,
但是满流 n*k 一定能推出有k种匹配方式吗？ 想不太清楚
唉 ~ 神奇的网络流建图！
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 210;
const int maxm = 1e6+10;
#define pb(x) push_back(x)
#define sc(x) scanf("%d",&x)
#define pfn(x) printf("%d\n",x)
#define pn printf("\n");
int T,n,m,fn;
int h[maxn];
int l[maxn];
int cur[maxn];
int tot;
struct edge{
int to,c,next;
edge(int to = 0, int c = 0, int next = 0) : to(to), c(c), next(next) {}
}es[maxm*2];
void add_edge(int u, int v, int c)
{
es[tot] = edge(v,c,h[u]);
h[u] = tot++;
}
bool bfs(int s, int t)
{
memset(l,0,sizeof(l));
l[s] = 1;
queue <int> q;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
//cout << u << endl;
if(u == t) return true;
for(int i = h[u]; i != -1; i = es[i].next)
{
int v = es[i].to;
if(!l[v] && es[i].c) {l[v] = l[u] + 1; q.push(v);}
}
}
return false;
}
int dfs(int x, int t, int mf)
{
if(x == t) return mf;
int ret = 0;
for(int &i = cur[x]; i != -1; i = es[i].next)
{
if(es[i].c && l[x] == l[es[i].to] - 1)
{
int f = dfs(es[i].to,t,min(mf,es[i].c));
es[i].c -= f;
es[i^1].c += f;
ret += f;
if(ret == mf) return ret;
}
}
return ret;
}
int dinic(int s, int t)
{
int res = 0;
while(bfs(s,t))
{
for(int i = 0; i <= 2*n+1; i++) cur[i] = h[i];
res += dfs(s,t,INF);
// //cout << "**************"<< res << endl;
}
return res;
}
bool g[maxn][maxn];
int f[maxn];
set <int> cnt[maxn];
void init()
{
for(int i = 0; i <= n; i++) f[i] = i;
}
int find(int x)
{
return f[x] == x ? x : f[x] = find(f[x]);
}
void unit(int x, int y)
{
int fx = find(x);
int fy = find(y);
f[fx] = fy;
}
bool same(int x, int y)
{
return find(x) == find(y);
}
bool check(int s, int t, int k)
{
memset(h,-1,sizeof(h));
tot = 0;
for(int i = 1; i <= n; i++)
{
add_edge(s,i,k);add_edge(i,s,0);
add_edge(n+i,t,k);add_edge(t,n+i,0);
}
for(int i = 1 ; i <= n; i++)
{
int fi = find(i);
for(int j = 1; j <= n; j++)
{
if(cnt[fi].count(j) == 1)
{
add_edge(i,n+j,1);
add_edge(n+j,i,0);
}
}
}
int res = dinic(s,t);
//cout <<"*" <<res << endl;
return res == n*k;
}
int main()
{
sc(T);
while(T--)
{
sc(n);sc(m);sc(fn);
init();
memset(g,false,sizeof(g));
for(int i = 0; i <= n; i++) cnt[i].clear();
for(int i = 1; i <= m; i++)
{
int a,b;
sc(a);sc(b);
g[a][b] = true;
}
for(int i = 0; i < fn; i++)
{
int a,b;
sc(a);sc(b);
unit(a,b);
}
for(int i = 1; i <= n; i++)
{
int fi = find(i);
for(int j = 1; j <= n; j++)
{
if(g[i][j]) cnt[fi].insert(j);
}
}
//for(int i = 1; i <= n; i++) //cout << cnt[i].size() << "&&&&"<<endl;
int s = 0,t = 2*n+1;
int st = 0,ed = n+1;
while(ed - st > 1)
{
int mid = st + (ed - st)/2;
//cout << st << " " << mid << " " << ed << endl;
if(check(s,t,mid)) st = mid;
else ed = mid;
}
//cout << "ans = ";
pfn(st);
}
return 0;
}