BZOJ 2424: [HAOI2010]订货

2424: [HAOI2010]订货

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 915  Solved: 639
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Description

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。

Input

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)

Output

只有1行，一个整数，代表最低成本

Sample Input

3 1 1000
2 4 8
1 2 4

Sample Output

34

HINT

 

Source

Day1

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动态规划。

一开始看到题面的时候以为是那道经典的贪心问题，就是维护当前最优单价，不断转移。

然而这道题限制了仓库的容量，使得最优单价不能直接向后转移，所以需要动态规划。

设$f[i][j]$表示经过了第$i$个月（此时已经到了月底），仓库中剩余量为$j$的当前最小费用。

转移如下：

$f[i][j+1]=min(f[i][j+1],f[i][j]+D_{i})$ 表示可以在本月额外购进一单位作为储存。

当$j \geq U_{i+1}$， $f[i+1][j-U_{i+1}]=min(f[i+1][j-U_{i+1}],f[i][j]+M*j$ 表示如果当前储备够下个月出售，下个月可以不购进商品，只需支付两个月之间的储存费用。

当$j \lt U_{i+1}$， $f[i+1][0]=min(f[i+1][0],f[i][j]+M*j+(U_{i+1}-j)*D_{i+1}$ 表示如果当前储备不够下个月，下个月除了需要支付储存费用，还需要额外购进一些商品。

 #include <cstdio>
 
 inline int nextChar(void)
 {
     const static int siz = ;
 
     static char buf[siz];
     static char *hd = buf + siz;
     static char *tl = buf + siz;
 
     if (hd == tl)
         fread(hd = buf, , siz, stdin);
 
     return *hd++;
 }
 
 inline int nextInt(void)
 {
     register int ret = ;
     register int neg = false;
     register int bit = nextChar();
 
     for (; bit < ; bit = nextChar())
         if (bit == '-')neg ^= true;
 
     for (; bit > ; bit = nextChar())
         ret = ret *  + bit - ;
 
     return neg ? -ret : ret;
 }
 
 const int inf = 1e9;
 const int maxn = ;
 const int maxm = ;
 
 int N, M, S;
 int U[maxn];
 int D[maxn];
 
 int f[maxn][maxm];
 
 inline void Min(int &a, const int &b)
 {
     if (a > b)a = b;
 }
 
 signed main(void)
 {
     N = nextInt();
     M = nextInt();
     S = nextInt();
 
     for (int i = ; i <= N; ++i)
         U[i] = nextInt();
 
     for (int i = ; i <= N; ++i)
         D[i] = nextInt();
 
     for (int i = ; i <= N; ++i)
         for (int j = ; j <= S; ++j)
             f[i][j] = inf;
 
     f[][] = D[] * U[];
 
     for (int i = ; i <= N; ++i)
         for (int j = ; j <= S; ++j)
             if (f[i][j] < inf)
             {
                 Min(f[i][j + ], f[i][j] + D[i]);
                 if (j >= U[i + ])
                     Min(f[i + ][j - U[i + ]], f[i][j] + j * M);
                 else
                     Min(f[i + ][], f[i][j] + (U[i + ] - j) * D[i + ] + j * M);
             }
 
     int ans = inf;
 
     for (int i = ; i <= S; ++i)
         Min(ans, f[N][i]);
 
     printf("%d\n", ans);
 }
  

突然，机房小伙伴们惊奇地看我：“这特么不是裸的费用流吗？”

WOC，我是有多蠢，去想DP？LTY神犇要来HACK我了，好怕怕~~~

补上最小费用流代码……

 #include <cstdio>
 #include <cstring>
 
 inline char nextChar(void)
 {
     static const int siz = ;
 
     static char buf[siz];
     static char *hd = buf + siz;
     static char *tl = buf + siz;
 
     if (hd == tl)
         fread(hd = buf, , siz, stdin);
 
     return *hd++;
 }
 
 inline int nextInt(void)
 {
     register int ret = ;
     register int neg = false;
     register int bit = nextChar();
 
     for (; bit < ; bit = nextChar())
         if (bit == '-')neg ^= true;
 
     for (; bit > ; bit = nextChar())
         ret = ret *  + bit - ;
 
     return neg ? -ret : ret;
 }
 
 const int siz = ;
 const int inf = ;
 
 int tot;
 int s, t;
 int hd[siz];
 int to[siz];
 int fl[siz];
 int vl[siz];
 int nt[siz];
 
 inline void add(int u, int v, int f, int w)
 {
     nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; vl[tot] = +w; hd[u] = tot++;
     nt[tot] = hd[v]; to[tot] = u; fl[tot] = ; vl[tot] = -w; hd[v] = tot++;
 }
 
 int dis[siz];
 int pre[siz];
 
 inline bool spfa(void)
 {
     static int que[siz];
     static int inq[siz];
     static int head, tail;
 
     memset(dis, 0x3f, sizeof(dis));
     memset(inq, , sizeof(inq));
     head = , tail = ;
     que[tail++] = s;
     pre[s] = -;
     dis[s] = ;
     inq[s] = ;
 
     while (head != tail)
     {
         int u = que[head++], v; inq[u] = ;
 
         for (int i = hd[u]; ~i; i = nt[i])
             if (dis[v = to[i]] > dis[u] + vl[i] && fl[i])
             {
                 dis[v] = dis[u] + vl[i], pre[v] = i ^ ;
                 if (!inq[v])inq[que[tail++] = v] = ;
             }
     }
 
     return dis[t] < 0x3f3f3f3f;
 }
 
 inline int minCost(void)
 {
     int cost = , flow;
 
     while (spfa())
     {
         flow = inf;
 
         for (int i = pre[t]; ~i; i = pre[to[i]])
             if (flow > fl[i ^ ])flow = fl[i ^ ];
 
         for (int i = pre[t]; ~i; i = pre[to[i]])
             fl[i] += flow, fl[i ^ ] -= flow;
 
         cost += dis[t] * flow;
     }
 
     return cost;
 }
 
 int n, m, k;
 
 signed main(void)
 {
     n = nextInt();
     m = nextInt();
     k = nextInt();
 
     s = , t = n + ;
 
     memset(hd, -, sizeof(hd));
 
     for (int i = ; i <= n; ++i)
         add(i, t, nextInt(), );
 
     for (int i = ; i <= n; ++i)
         add(s, i, inf, nextInt());
 
     for (int i = ; i < n; ++i)
         add(i, i + , k, m);
 
     printf("%d\n", minCost());
 }

@Author: YouSiki