hdu 3718

一个二分图最大匹配的题；

匈牙利算法不熟；

建了个模，用最小费用最大流解决了

 #include <iostream>
 #include <cstring>
 #define INF 9999999
 #include <cstdio>
 #include <queue>
 #include <vector>
 #include<algorithm>
 using namespace std;
 #define maxn 6100

 struct  edge
 {
     int from,to,cap,flow,cost;
 };
 struct MCMF
 {
     int n,m,s,t;
     vector<edge>edges;
     vector<int>G[maxn];
     int inq[maxn];
     int d[maxn];
     int p[maxn];
     int a[maxn];
     void init(int n)
     {
         this->n=n;
         for(int i=; i<n; i++)
             G[i].clear();
         edges.clear();
     }
     void addedge(int from,int to,int cap,int cost)
     {
         edges.push_back((edge){from,to,cap,,cost});
         edges.push_back((edge){to,from,,,-cost});
         m=edges.size();
         G[from].push_back(m-);
         G[to].push_back(m-);
     }

     bool bellman(int s,int t,int &flow,int &cost)
     {
         for(int i=; i<n; i++)d[i]=INF;
         memset(inq,,sizeof(inq));
         d[s]=;
         inq[s]=;
         p[s]=;
         a[s]=INF;

         queue<int>Q;
         Q.push(s);
         while(!Q.empty())
         {
             int u = Q.front();
             Q.pop();
             inq[u] = ;
             for(int i = ; i < G[u].size(); i++)
             {
                 edge& e = edges[G[u][i]];
                 if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
                 {
                     d[e.to] = d[u] + e.cost;
                     p[e.to] = G[u][i];
                     a[e.to] = min(a[u], e.cap - e.flow);
                     if(!inq[e.to])
                     {
                         Q.push(e.to);
                         inq[e.to] = ;
                     }
                 }
             }
         }
         if(d[t] == INF) return false;
         flow += a[t];
         cost += d[t]*a[t];
         int u = t;
         while(u != s)
         {
             edges[p[u]].flow += a[t];
             edges[p[u]^].flow -= a[t];
             u = edges[p[u]].from;
         }
         return true;
     }
     int Mincost(int s, int t)
     {
         int flow = , cost = ;
         while(bellman(s, t, flow, cost));
         return cost;
     }
 };
 int n;
 int k,m;
 char s2[],s1[];
 int cur[][];
 int tot[];
 void first_solve()
 {
     memset(cur,,sizeof(cur));
     memset(tot,,sizeof(tot));
     for(int i=; i<=n; i++)
     {
         int k1=s1[i]-'A'+;
         int k2=s2[i]-'A'+;
         tot[k1]++;
         cur[k1][k2]++;
     }
 }
 MCMF solve;
 int main()
 {
     int t;
     int st=;
     int final=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d%d",&n,&k,&m);
         for(int i=; i<=n; i++)
         {
             char s[];
             scanf("%s",s);
             s1[i]=s[];
         }
         for(int d=; d<=m; d++)
         {
             solve.init(final+);
             for(int j=; j<=n; j++)
             {
                 char s[];
                 scanf("%s",s);
                 s2[j]=s[];
             }
             first_solve();
             for(int i=; i<=; i++)
                 solve.addedge(st,i,,);
             for(int i=; i<=; i++)
             {
                 for(int j=; j<=; j++)
                 {
                     int cnt=+j;
                     if (cur[i][j])
                         solve.addedge(i,cnt,,tot[i]-cur[i][j]);
                     else solve.addedge(i,cnt,,tot[i]);
                 }
             }
             for(int i=;i<=;i++)solve.addedge(i+,final,,);
             int ans=n-solve.Mincost(st,final);
             printf("%.4lf\n",(double)ans/(double)n);
         }
     }
     return ;
 }