hdu3081 Marriage Match II(最大流)

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Marriage Match II

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2410????Accepted Submission(s): 820

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

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題意：

有n個女孩和n個男孩，有m個女孩與男孩的關系，代表女孩喜歡男孩，有f個女孩與女孩的關系，代表女孩與女孩是好朋友。

若女孩i與女孩j是好朋友，而女孩i喜歡男孩k，則女孩j也可以和男孩匹配。即女孩之間的關系是可以傳遞的，而男孩之間的不可以。

每對男孩與女孩之間只能匹配一次，問可以進行幾輪配對。

分析：
先floyd搞好女孩之間的傳遞關系，然后對于可以配對的女孩與男孩之間連一條容量為1的邊，然后二分答案，每次二分之后從源點向各個女孩連容量為二分值的邊，從各個男孩向匯點連容量為二分值的邊。

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
#define MAXN 1010
struct edge{
    int to,cap,rev;
    edge(int _to,int _cap,int _rev)
    {
        to=_to;
        cap=_cap;
        rev=_rev;
    }
};
const int MAX_V=5020;
vector<edge>G[MAX_V];
int iter[MAX_V];
int head[MAXN];
int _to[510*510];
int _flow[510*510];
int _next[510*510];
int level[MAX_V];
int tot=0;
void add_edge(int from,int to,int cap)
{
    G[from].PB(edge(to,cap,G[to].size()));
    G[to].PB(edge(from,0,G[from].size()-1));
}
void Add(int u,int v,int f){
    _to[tot]=v;
    _flow[tot]=f;
    _next[tot]=head[u];
    head[u]=tot++;
}
void bfs(int s,int t)
{
    CLR(level,-1);
    queue<int>q;
    level[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=0;i<G[u].size();i++)
        {
            edge &e=G[u][i];
            if(e.cap>0&&level[e.to]<0)
            {
                level[e.to]=level[u]+1;
                q.push(e.to);
            }
        }
    }
}
int dfs(int v,int t,int f)
{
    if(v==t)return f;
    for(int &i=iter[v];i<G[v].size();i++)
    {
        edge &e=G[v][i];
        if(e.cap>0&&level[v]<level[e.to])
        {
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0)
            {
                e.cap-=d;;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}
int Dinic(int s,int t)
{
    int flow=0;
    for(;;)
    {
        bfs(s,t);
        if(level[t]<0)return flow;
        memset(iter,0,sizeof(iter));
        int f;
        while((f=dfs(s,t,INF))>0)
        {
            flow+=f;
        }
    }
}

int a[210][210];

int main()
{
    ios::sync_with_stdio(false);
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m,f;
        scanf("%d%d%d",&n,&m,&f);
        tot=0;
        for(int i=0;i<MAXN;i++)head[i]=-1;
        int u,v;
        memset(a,0,sizeof(a));
        for(int i=0;i<m;i++){
            scanf("%d%d",&u,&v);
            a[u][v+n]=1;
        }
        for(int i=0;i<f;i++){
            scanf("%d%d",&u,&v);
            a[u][v]=1;
            a[v][u]=1;
        }
        for(int i=1;i<=n+n;i++)a[i][i]=1;
        for(int k=1;k<=n+n;k++){
            for(int i=1;i<=n+n;i++){
                for(int j=1;j<=n+n;j++){
                    a[i][j]=max(a[i][j],a[i][k]&a[k][j]);
                }
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1+n;j<=n+n;j++){
                if(a[i][j])Add(i,j,1);
            }
        }
        int l=0,r=n;
        int s=0,t=2*n+1;
        int ans=0;
        while(l<=r){
            int mid=(l+r)>>1;
            for(int i=0;i<=2*n+1;i++)G[i].clear();
            for(int i=1;i<=2*n;i++){
                int now=head[i];
                while(now!=-1){
                    add_edge(i,_to[now],_flow[now]);
                    now=_next[now];
                }
            }
            for(int i=1;i<=n;i++){
                add_edge(s,i,mid);
                add_edge(n+i,t,mid);
            }
            if(Dinic(s,t)==mid*n){
                ans=mid;
                l=mid+1;
            }
            else r=mid-1;
        }
        printf("%d\n",ans);
    }
        
                    
            
    return 0;
}

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