思路

神奇的网络流优化
这题相当于修车的数据加强版XD
依然是用点表示第i个厨师倒数第j个做某个菜，贡献就是会使之后的人都加上这个菜的等待时间
然后这样有60pts
考虑优化，发现每次只可能是某一个厨师做一道菜，所以每次只给上次做菜的那个厨师加一层边就好了
然后我常数太大只能在luogu上吸氧

代码

// luogu-judger-enable-o2
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 700000;
const int INF = 0x3f3f3f3f;
struct Edge{
    int u,v,cap,cost,flow;
};
vector<Edge> edges;
vector<int> G[MAXN];
void addedge(int u,int v,int cap,int cost){
    edges.push_back((Edge){u,v,cap,cost,0});
    edges.push_back((Edge){v,u,0,-cost,0});
    int cnt=edges.size();
    G[u].push_back(cnt-2);
    G[v].push_back(cnt-1);
}
queue<int> q;
int s,t,a[MAXN],d[MAXN],p[MAXN],vis[MAXN];
bool spfa(int &flow,int &cost){
    memset(d,0x3f,sizeof(d));
    memset(p,0,sizeof(p));
    q.push(s);
    vis[s]=true;
    a[s]=INF;
    d[s]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        vis[x]=false;
        for(int i=0;i<G[x].size();i++){
            Edge &e = edges[G[x][i]];
            if(e.cap>e.flow&&d[x]+e.cost<d[e.v]){
                d[e.v]=d[x]+e.cost;
                a[e.v]=min(a[x],e.cap-e.flow);
                p[e.v]=G[x][i];
                if(!vis[e.v]){
                    vis[e.v]=true;
                    q.push(e.v);
                }
            }
        }
    }
    if(d[t]==INF)
        return false;
    flow+=a[t];
    cost+=d[t]*a[t];
    for(int i=t;i!=s;i=edges[p[i]].u){
        edges[p[i]].flow+=a[t];
        edges[p[i]^1].flow-=a[t];
    }
    return true;
}
int n,m,sum,val[50][120],who[MAXN],num[MAXN];
void MCMF(int &cost,int &flow){
    cost=0,flow=0;
    while(spfa(flow,cost)){
        int mid=edges[p[t]].u;
        int times=num[mid-1];
        int tmid=who[mid];
        addedge(mid-1,t,1,0);
        for(int i=1;i<=n;i++)
            addedge(i,mid-1,INF,(sum-times+1)*val[i][tmid]);
    }
}
int main(){
    //freopen("testdata.in","r",stdin);
    scanf("%d %d",&n,&m);
    s=MAXN-2;
    t=MAXN-3;
    for(int i=1;i<=n;i++){
        int mid;
        scanf("%d",&mid);
        sum+=mid;
        addedge(s,i,mid,0);
        who[i]=i;
    }
    for(int i=1;i<=m;i++){
      for(int j=1;j<=sum;j++){
            who[(i-1)*sum+j+n]=i;
            num[(i-1)*sum+j+n]=j;
      }
    }
    for(int i=1;i<=m;i++)
        addedge(i*sum+n,t,1,0);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&val[i][j]);
            addedge(i,j*sum+n,INF,val[i][j]);
        }
    }
    //printf("ok\n");
    int cost,flow;
    MCMF(cost,flow);
    //printf("ok\n");
    printf("%d\n",cost);
    return 0;
}