poj3686

题解：

KM算法

把每一个点拆成n个

然后改变编圈

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=105;
char s[5],q[N][5];
int cnt,a[N][N],z,e,cas,b[N*N];
int visr[N],T,x,y,exl[N],exr[N],visl[N],match[N],slack[N],n,m;
int in(char s[])
{
    for (int i=1;i<=cnt;i++)
     if (q[i][0]==s[0])return i;
    q[++cnt][0]=s[0]; 
    return cnt; 
}
int dfs(int x)
{
    visl[x]=1;
    for (int i=1;i<=m;i++)
     if (!visr[i])
      {
          int k=exl[x]+exr[i]-a[x][i];
          if (k==0)
           {
               visr[i]=1;
               if (!match[i]||dfs(match[i]))
                {
                    match[i]=x;
                    return 1;
                }
           }
          else slack[i]=min(slack[i],k); 
      }
    return 0;  
}
int km()
{
    memset(exl,0,sizeof exl);
    memset(exr,0,sizeof exr);
    for (int i=1;i<=m;i++)
     for (int j=1;j<=m;j++)exl[i]=max(exl[i],a[i][j]);
    memset(match,0,sizeof match); 
    for (int i=1;i<=m;i++)
     {
        memset(slack,0x3f,sizeof slack);
        while (1)
         {
              memset(visl,0,sizeof visl);
              memset(visr,0,sizeof visr);
              if (dfs(i))break;
              int d=1e9;
              for (int j=1;j<=m;j++)
               if (!visr[j])d=min(d,slack[j]);
              for (int j=1;j<=m;j++)
             if (visl[j])exl[j]-=d;
            for (int j=1;j<=m;j++)
             if (visr[j])exr[j]+=d;
             else slack[j]-=d;
         }
     } 
    int ans=0,flag=0; 
    for (int i=1;i<=m;i++)ans+=a[match[i]][i];
    return ans;
}
int main()
{
    scanf("%d",&T);
    while (T--)
     {
         scanf("%d%d%d",&n,&m,&e);
         cnt=0;
         memset(q,0,sizeof q);
         for (int i=1;i<=n;i++)
          {
              scanf("%s",s);
              b[i]=in(s);
          } 
         while (e--)
         {
             cnt=0;
             memset(q,0,sizeof q);
             memset(a,0,sizeof a);
              for (int i=1;i<=n;i++)
               {
                   scanf("%s",s);
                   a[b[i]][in(s)]++;
               }
              printf("%.4lf
",(double)km()/n); 
         } 
     }
}