这道题就是“修车”的数据加强版……但是数据范围扩大了好多，应对方法是“动态开点”。
首先先把“所有厨师做的倒数第一道菜”和所有菜连边，然后跑一下spfa，找出哪一个厨师在增广路上，把“这个厨师做的倒数第二道菜”和所有菜连边，然后继续spfa，如此循环往复直到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;
}