G_M_C_美食节

美食节

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 <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define em emplace
#define pii pair<int,int>
#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 second
using 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<int> 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;
}