某公司估计市场在第i个月对某产品的需求量为Ui，已知在第i月该产品的订货单价为di，上个月月底未销完的单位产品要付存贮费用m，假定第一月月初的库存量为零，第n月月底的库存量也为零，问如何安排这n个月订购计划，才能使成本最低？每月月初订购，订购后产品立即到货，进库并供应市场，于当月被售掉则不必付存贮费。假设仓库容量为S。

第1行：n, m, S (0<=n<=50, 0<=m<=10, 0<=S<=10000)

第2行：U1 , U2 , ... , Ui , ... , Un (0<=Ui<=10000)

第3行：d1 , d2 , ..., di , ... , dn (0<=di<=100)

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

queue<int> q;

int head[60] , to[500] , val[500] , cost[500] , next[500] , cnt = 1 , s , t , dis[60] , from[60] , pre[60];

void add(int x , int y , int v , int c)

{

	to[++cnt] = y , val[cnt] = v , cost[cnt] = c , next[cnt] = head[x] , head[x] = cnt;

	to[++cnt] = x , val[cnt] = 0 , cost[cnt] = -c , next[cnt] = head[y] , head[y] = cnt;

}

bool spfa()

{

	int x , i;

	memset(from , -1 , sizeof(from));

	memset(dis , 0x3f , sizeof(dis));

	dis[s] = 0 , q.push(s);

	while(!q.empty())

	{

		x = q.front() , q.pop();

		for(i = head[x] ; i ; i = next[i])

			if(val[i] && dis[to[i]] > dis[x] + cost[i])

				dis[to[i]] = dis[x] + cost[i] , from[to[i]] = x , pre[to[i]] = i , q.push(to[i]);

	}

	return ~from[t];

}

int mincost()

{

	int ans = 0 , i , k;

	while(spfa())

	{

		k = 0x3f3f3f3f;

		for(i = t ; i != s ; i = from[i]) k = min(k , val[pre[i]]);

		ans += k * dis[t];

		for(i = t ; i != s ; i = from[i]) val[pre[i]] -= k , val[pre[i] ^ 1] += k;

	}

	return ans;

}

int main()

{

	int n , m , k , i , x;

	scanf("%d%d%d" , &n , &m , &k) , s = 0 , t = n + 1;

	for(i = 1 ; i <= n ; i ++ ) scanf("%d" , &x) , add(i , t , x , 0);

	for(i = 1 ; i <= n ; i ++ ) scanf("%d" , &x) , add(s , i , 0x3f3f3f3f , x);

	for(i = 1 ; i < n ; i ++ ) add(i , i + 1 , k , m);

	printf("%d\n" , mincost());

	return 0;

}