HDU 3081 Marriage Match II

Marriage Match II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3837    Accepted Submission(s): 1255

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

 

二分求轮数，Dinic算法模板

  1 #include <iostream>  2 #include <cstring>  3 #include <vector>  4 #include <queue>  5 #define maxn 210  6 #define INF 0X3fffffff   7 using namespace std;  8 struct Edge    9 {   10     int from,to,cap,flow;   11     Edge(){}   12     Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}   13 };   14    15 struct Dinic   16 {   17     int n,m,s,t;   18     vector<Edge> edges;   19     vector<int> G[maxn];   20     bool vis[maxn];   21     int cur[maxn];   22     int d[maxn];   23    24     void init(int n,int s,int t)   25     {   26         this->n=n, this->s=s, this->t=t;   27         edges.clear();   28         for(int i=0;i<=n;++i) G[i].clear();   29     }   30    31     void AddEdge(int from,int to,int cap)   32     {   33         edges.push_back( Edge(from,to,cap,0) );   34         edges.push_back( Edge(to,from,0,0) );   35         m=edges.size();   36         G[from].push_back(m-2);   37         G[to].push_back(m-1);   38     }   39    40     bool BFS()   41     {   42         queue<int> Q;   43         memset(vis,0,sizeof(vis));   44         vis[s]=true;   45         d[s]=0;   46         Q.push(s);   47         while(!Q.empty())   48         {   49             int x=Q.front(); Q.pop();   50             for(int i=0;i<G[x].size();++i)   51             {   52                 Edge& e=edges[G[x][i]];   53                 if(!vis[e.to] && e.cap>e.flow)   54                 {   55                     vis[e.to]=true;   56                     d[e.to]=d[x]+1;   57                     Q.push(e.to);   58                 }   59             }   60         }   61         return vis[t];   62     }   63    64     int DFS(int x,int a)   65     {   66         if(x==t || a==0) return a;   67         int flow=0, f;   68         for(int &i=cur[x];i<G[x].size();++i)   69         {   70             Edge &e=edges[G[x][i]];   71             if(d[e.to]==d[x]+1 && (f=DFS(e.to,min(a,e.cap-e.flow) ) )>0)   72             {   73                 e.flow +=f;   74                 edges[G[x][i]^1].flow -=f;   75                 flow +=f;   76                 a -=f;   77                 if(a==0) break;   78             }   79         }   80         return flow;   81     }   82    83     int max_flow()   84     {   85         int ans=0;   86         while(BFS())   87         {   88             memset(cur,0,sizeof(cur));   89             ans +=DFS(s,INF);   90         }   91         return ans;   92     }   93 }DC; 94 int parnet[maxn]; 95 bool flag[maxn][maxn]; 96 int find(int n){ 97     if(n==parnet[n]){ 98         return n; 99     }100     else return parnet[n]=find(parnet[n]);101 }102 void Union(int x,int y){103     int xx=find(x);104     int yy=find(y);105     if(xx!=yy){106         parnet[xx]=yy;107     }108 }109 bool solve(int n,int limit){110     int start=0,end=2*n+1;111     DC.init(n*2+2,start,end);112     for(int i=1;i<=n;i++){113         for(int j=1;j<=n;j++){114             if(flag[i][j]){115                 DC.AddEdge(i,j+n,1);116             }117         }118     }119     for(int i=1;i<=n;i++){120         DC.AddEdge(i+n,end,limit);121     }122     for(int i=1;i<=n;i++){123         DC.AddEdge(start,i,limit);124     }125     return DC.max_flow()==n*limit;126 }127 void init(int n){128     for(int i=1;i<=n;i++){129         parnet[i]=i;130     }131     memset(flag,0,sizeof(flag));132 }133 int main(){134     int n,m,f,T,a,b;135     cin.sync_with_stdio(false);136     cin>>T;137     while(T--){138         cin>>n>>m>>f;139         init(n);140         while(m--){141             cin>>a>>b;142             flag[a][b]=true;143         }144         while(f--){145             cin>>a>>b;146             Union(a,b);147         }148         for(int i=1;i<=n;i++){149             for(int j=i+1;j<=n;j++){150                 if(find(i)==find(j)){151                     for(int k=1;k<=n;k++){152                         flag[i][k]=flag[j][k]=(flag[i][k]||flag[j][k]);153                     }154                 }155             }156         }157         int l=0,r=n;158         while(l<r){159             int mid=(l+r+1)/2;160             if(solve(n,mid)){161                 l=mid;162             }163             else{164                 r=mid-1;165             }166         }167         cout<<l<<endl;168     }169     return 0;170 }

 

2017-03-06 12:12:51