题目链接：https://www.nowcoder.com/acm/contest/143/E
思路：要让搬寝室的人最少，也就是要让住在原寝室的人最多，所以要确定的就是现在的寝室和原来哪个寝室匹配，只要跑个最大权匹配就行了，km模板。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int g[108][108],cy[108],cx[108],qw[108],slack[108],n,a[108][4],b[108][4];
bool visx[108],visy[108];
bool dfs(int u)
{
visx[u]=1;
FOR(i,1,n)
{
if(visy[i]) continue;
int tmp=cx[u]+cy[i]-g[u][i];
if(tmp==0)
{
visy[i]=1;
if(qw[i]==-1||dfs(qw[i]))
{
qw[i]=u;
return 1;
}
}
else slack[i]=min(slack[i],tmp);
}
return 0;
}
int km()
{
REW(qw,-1),REW(cy,0);
FOR(i,1,n)
{
cx[i]=-inf;
FOR(j,1,n) cx[i]=max(cx[i],g[i][j]);
}
FOR(i,1,n)
{
REW(slack,inf);
while(1)
{
REW(visx,0),REW(visy,0);
if(dfs(i)) break;
int d=inf;
FOR(j,1,n) if(!visy[j]) d=min(d,slack[j]);
FOR(j,1,n)
{
if(visx[j]) cx[j]-=d;
if(visy[j]) cy[j]+=d;
else slack[j]-=d;
}
}
}
int ans=0;
FOR(i,1,n) ans+=g[qw[i]][i];
return ans;
}
int main()
{
cin.tie(0);
cout.tie(0);
si(n);
FOR(i,1,n)
FOR(j,0,3) si(a[i][j]);
FOR(i,1,n)
FOR(j,0,3) si(b[i][j]);
FOR(i,1,n)
{
FOR(j,1,n)
{
int s=0;
FOR(k,0,3)
{
FOR(z,0,3)
{
if(a[i][k]==b[j][z]) s++;
}
}
g[i][j]=s;
}
}
cout<<4*n-km()<<endl;
return 0;
}