room
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
Nowcoder University has 4n students and n dormitories ( Four students per dormitory). Students numbered from 1 to 4n.

And in the first year, the i-th dormitory ‘s students are (x1[i],x2[i],x3[i],x4[i]), now in the second year, Students need to decide who to live with.

In the second year, you get n tables such as (y1,y2,y3,y4) denote these four students want to live together.

Now you need to decide which dormitory everyone lives in to minimize the number of students who change dormitory.

輸入描述:
The first line has one integer n.

Then there are n lines, each line has four integers (x1,x2,x3,x4) denote these four students live together in the first year

Then there are n lines, each line has four integers (y1,y2,y3,y4) denote these four students want to live together in the second year
輸出描述:
Output the least number of students need to change dormitory.
示例1
輸入
複製
2
1 2 3 4
5 6 7 8
4 6 7 8
1 2 3 5
輸出
複製
2
說明
Just swap 4 and 5
備註:
1<=n<=100

1<=x1,x2,x3,x4,y1,y2,y3,y4<=4n

It’s guaranteed that no student will live in more than one dormitories.

假設新的宿舍裡，第 i 個4人團體安排到第 j 間宿舍，那麼不用搬的學生數量，
就是4人團體和原住民的交，那麼費用就是用搬的學生數量，
變成帶權二分圖匹配問題，可以用費用流做（直接構完圖用最小費用最大流模板求解）

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 40010;

typedef struct Node{
    int s,to,next,capacity,value;
}Node;

int n,m;
int ecnt;
Node node[MAXM];
int pre[MAXN];
int head[MAXN];
int dis[MAXN];
bool vis[MAXN];

void init()
{
    ecnt = 0;
    memset(head,-1,sizeof(head));
    memset(node,0,sizeof(node));
}

void addEdge(int u,int v,int w,int c)
{
    node[ecnt].to = v;
    node[ecnt].s = u;
    node[ecnt].capacity = c;//流量
    node[ecnt].value = w;//路徑長度是費用
    node[ecnt].next = head[u];
    head[u] = ecnt++;
    node[ecnt].to = u;
    node[ecnt].s = v;
    node[ecnt].capacity = 0;//流量
    node[ecnt].value = -w;//路徑長度是費用
    node[ecnt].next = head[v];
    head[v] = ecnt++;
}

bool spfa(int s,int t,int nnum)
{
    memset(vis,0,sizeof(vis));
    memset(pre,-1,sizeof(pre));
    for(int i = 0;i <= nnum;++i)
    {
        dis[i] = inf;
    }
    queue<int>que;
    que.push(s);
    dis[s] = 0;
    vis[s] = true;
    while(!que.empty())
    {
        int temp = que.front();
        que.pop();
        vis[temp] = false;
        for(int i = head[temp];i + 1;i = node[i].next)
        {
            if(node[i].capacity)
            {
                int ne = node[i].to;
                if(dis[temp] + node[i].value < dis[ne]){
                    dis[ne] = dis[temp] + node[i].value;
                    pre[ne] = i;
                    if(!vis[ne]){
                        que.push(ne);
                        vis[ne] = true;
                    }
                }
            }
        }
    }
    if(dis[t] == inf)
        return false;
    return true;
}

int getMincost(int s,int t,int nnum)
{
    int ans_flow = 0;
    int ans_cost = 0;
    int temp,minc;
    while(spfa(s,t,nnum))
    {
//        cout << 0 << endl;
        temp = t;
        minc = inf;
        while(pre[temp] != -1)
        {
            minc = min(node[pre[temp]].capacity,minc);
            temp = node[pre[temp]].s;
        }
        temp = t;
        while(pre[temp] != -1)
        {
            node[pre[temp]].capacity -= minc;
            int ss = pre[temp] ^ 1;
            node[ss].capacity += minc;
            temp = node[pre[temp]].s;
        }
        ans_cost += dis[t] * minc;
    }
    return ans_cost;
}

typedef struct room{
    int num[4];
}room;
room a[MAXN],b[MAXN];

int main()
{
    while(~scanf("%d",&n))
    {
        init();
        for(int i = 0;i < n;++i)
        {
            for(int j = 0;j < 4;++j)
            {
                scanf("%d",&a[i].num[j]);
            }
        }
        for(int i = 0;i < n;++i)
        {
            for(int j = 0;j < 4;++j)
            {
                scanf("%d",&b[i].num[j]);
            }
        }
        for(int i = 0;i < n;++i)
        {
            for(int j = 0;j < n;++j)
            {
                int v = 0;
                for(int k = 0; k < 4;++k)
                {
                    for(int p = 0;p < 4;++p)
                    {
                        if(a[i].num[k] == b[j].num[p])
                            v++;
                    }
                }
                //第 i 個4人團體安排到第 j 間宿舍,需要搬的學生數量
                //只能搬一次，所以邊的容量為1
                addEdge(i + 1,n + j + 1,4 - v,1);
            }
        }
        //新增源點和匯點
        //與源點和匯點連線的邊費用為0
        for(int i = 0;i < n;++i)
        {
            addEdge(0,i + 1,0,1);
            addEdge(n + i + 1,2 * n + 1,0,1);
        }
        int ans;
        printf("%d\n",getMincost(0,2 * n + 1,2 * n + 1));
    }
    return 0;
}