這道題就是“修車”的數據加強版……但是數據範圍擴大了好多，應對方法是“動態開點”。

首先先把“所有廚師做的倒數第一道菜”和所有菜連邊，然後跑一下spfa，找出哪一個廚師在增廣路上，把“這個廚師做的倒數第二道菜”和所有菜連邊，然後繼續spfa，如此循環往復直到spfa找不出最短路。

#include <queue>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
 

#define enter putchar('\n')
typedef long long ll;
using namespace std;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0'
 
 && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 100005, M = 10000005, INF = 0x3f3f3f3f;
int n, m, src, des, tot, ans, sum;
int
 
 cook[105][805], dish[805], id[N], rnk[N], w[45][105], p[45];
int ecnt = 1, adj[N], dis[N], pre[N], nxt[M], go[M], cap[M], cost[M];

void _add(int u, int v, int w, int c){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
    cap[ecnt] = w;
    cost[ecnt] = c;
}
void add(int u, int v, int w, int c){
    _add(u, v, w, c);
    _add(v, u, 0, -c);
}
bool spfa(){
    queue <int> que;
    static bool inq[N] = {0};
    for(int i = 1; i <= tot; i++)
        dis[i] = INF, pre[i] = 0;
    dis[src] = 0, que.push(src), inq[src] = 1;
    while(!que.empty()){
        int u = que.front();
        que.pop(), inq[u] = 0;
        for(int e = adj[u], v; e; e = nxt[e]){
            if(cap[e] && dis[u] + cost[e] < dis[v = go[e]]){
                dis[v] = dis[u] + cost[e], pre[v] = e;
                if(!inq[v]) que.push(v), inq[v] = 1;
            }
        }
    }
    return dis[des] < INF;
}

int main(){

    read(n), read(m);
    src = ++tot, des = ++tot;
    for(int i = 1; i <= n; i++)
        read(p[i]), sum += p[i];
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            read(w[i][j]);
    for(int i = 1; i <= n; i++){
        dish[i] = ++tot, id[tot] = i;
        add(src, dish[i], p[i], 0);
    }
    for(int i = 1; i <= m; i++){
        cook[i][1] = ++tot, id[tot] = i, rnk[tot] = 1;
        add(cook[i][1], des, 1, 0);
        for(int j = 1; j <= n; j++)
            add(dish[j], cook[i][1], 1, w[j][i]);
    }
    while(spfa()){
        int tmp = go[pre[des] ^ 1], _cook = id[tmp], _rank = rnk[tmp], flow = INF;
        for(int e = pre[des]; e; e = pre[go[e ^ 1]])
            flow = min(flow, cap[e]);
        for(int e = pre[des]; e; e = pre[go[e ^ 1]])
            cap[e] -= flow, cap[e ^ 1] += flow;
        ans += flow * dis[des];
        cook[_cook][_rank + 1] = ++tot, id[tot] = _cook, rnk[tot] = _rank + 1;
        add(tot, des, 1, 0);
        for(int i = 1; i <= n; i++)
            add(dish[i], tot, 1, w[i][_cook] * (_rank + 1));
    }
    write(ans), enter;

    return 0;
}

BZOJ 2879 [Noi2012]美食節 | 費用流 動態開點