题解：学习了动态加边，可以说是进一步理解了网络流。具体思路就是考虑每一道菜，如果这是该位厨师最后一次做，那么等待时间就是做这道菜的时间，如果是倒数第二次做，就要两倍时间（目前做了一次，后面还有等待的一次时间）……，对于其他菜以此类推。那么可以这样考虑，当这位厨师倒一次做的边没有流量的时候，是不可能用到倒二次做的边（因为费用流会优先选择费用小的边，而我们的时间花费即为费用，故倒一次没用不可能使用倒二次），于是乎可以在跑完每次最短路的时候，判断这次是从哪个厨师的哪个点转移来的，紧接着连上接下来的一个点（倒一就连倒二，倒二就倒三……）。

#pragma comment(linker, "/STACK: 1024000000,1024000000")#include 
    
     #define pb push_back#define mp make_pair#define eb emplace_back#define em emplace#define pii pair
     
      #define de(x) cout << #x << " = " << x << endl#define clr(a,b) memset(a,b,sizeof(a))#define INF (0x3f3f3f3f)#define LINF ((long long)(0x3f3f3f3f3f3f3f3f))#define F first#define S secondusing namespace std;inline int getint(){  int _x=0; char _tc=getchar();  while(_tc<'0'||_tc>'9') _tc=getchar();  while(_tc>='0'&&_tc<='9') _x*=10,_x+=(_tc-'0'),_tc=getchar();  return _x;}const int M = 1e5 + 15;const int N = 110;int T[N][N];int p[N], P;int n, m, st, ed;struct Edge{    int u, v, c, w, nxt;    Edge(){}    Edge( int t2, int t3, int t4, int t5, int t6 ) :        u(t2), v(t3), c(t4), w(t5), nxt(t6) {}};Edge e[M<<4];int ect;int pr[M], h[M], d[M];bool vis[M];int ans = 0;void _add( int u, int v, int c, int w ){    e[ect].u = u; e[ect].v = v; e[ect].c = c; e[ect].w = w; e[ect].nxt = h[u]; h[u] = ect ++;    e[ect].u = v; e[ect].v = u; e[ect].c = 0; e[ect].w = -w; e[ect].nxt = h[v]; h[v] = ect ++;}void init(){    clr(h,-1);    P = ect = 0;}bool spfa(){    clr(d,INF);    clr(vis,false);    clr(pr,-1);    queue
      
        Q;    Q.push( st );    d[st] = 0;    vis[st] = true;    while ( !Q.empty() )    {        int u = Q.front(); Q.pop();        vis[u] = false;        for ( int i = h[u]; i + 1; i = e[i].nxt )        {            int v = e[i].v;            if ( e[i].c > 0 && d[v] > d[u] + e[i].w )            {                d[v] = d[u] + e[i].w;                pr[v] = i;                if ( !vis[v] )                {                    vis[v] = true;                    Q.push(v);                }            }        }    }    return d[ed] != INF;}void mcf(){    int f = INF;    for ( int i = ed; i != st; i = e[pr[i]].u )        f = min( f, e[pr[i]].c );    for ( int i = ed; i != st; i = e[pr[i]].u )    {        e[pr[i]].c -= f, e[pr[i]^1].c += f;        ans += f*e[pr[i]].w;    }    int x = e[pr[ed]].u;    int nw = (x-n-1)/P + 1, th = (x-n)%P + 1;    if ( th <= P )    {        for ( int i = 1; i <= n; i ++ )            _add( i, n+(nw-1)*P+th, 1, th*T[i][nw] );    }    _add( n+(nw-1)*P+th, ed, 1, 0 );}int main(){    init();    scanf("%d%d", &n, &m);    for ( int i = 1; i <= n; i ++ )        scanf("%d", &p[i]), P += p[i];    for ( int i = 1; i <= n; i ++ )        for ( int j = 1; j <= m; j ++ )            scanf("%d", &T[i][j]);    st = 0, ed = n + 1 + P*m;    for ( int i = 1; i <= n; i ++ )    {        _add( st, i, p[i], 0 );        for ( int j = 1; j <= m; j ++ )            _add( i, n+(j-1)*P+1, 1, T[i][j] );    }    for ( int i = 1; i <= m; i ++ )        _add( n+(i-1)*P+1, ed, 1, 0 );    while ( spfa() ) mcf();    printf("%d\n", ans);    return 0;}