2424: [HAOI2010]订货
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 915 Solved: 639
[Submit][Status][Discuss]
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
2 4 8
1 2 4
Sample Output
34
HINT
Source
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。
1 #include <cstdio> 2 3 inline int nextChar(void) 4 { 5 const static int siz = 1024; 6 7 static char buf[siz]; 8 static char *hd = buf + siz; 9 static char *tl = buf + siz; 10 11 if (hd == tl) 12 fread(hd = buf, 1, siz, stdin); 13 14 return *hd++; 15 } 16 17 inline int nextInt(void) 18 { 19 register int ret = 0; 20 register int neg = false; 21 register int bit = nextChar(); 22 23 for (; bit < 48; bit = nextChar()) 24 if (bit == '-')neg ^= true; 25 26 for (; bit > 47; bit = nextChar()) 27 ret = ret * 10 + bit - 48; 28 29 return neg ? -ret : ret; 30 } 31 32 const int inf = 1e9; 33 const int maxn = 55; 34 const int maxm = 10005; 35 36 int N, M, S; 37 int U[maxn]; 38 int D[maxn]; 39 40 int f[maxn][maxm]; 41 42 inline void Min(int &a, const int &b) 43 { 44 if (a > b)a = b; 45 } 46 47 signed main(void) 48 { 49 N = nextInt(); 50 M = nextInt(); 51 S = nextInt(); 52 53 for (int i = 1; i <= N; ++i) 54 U[i] = nextInt(); 55 56 for (int i = 1; i <= N; ++i) 57 D[i] = nextInt(); 58 59 for (int i = 0; i <= N; ++i) 60 for (int j = 0; j <= S; ++j) 61 f[i][j] = inf; 62 63 f[1][0] = D[1] * U[1]; 64 65 for (int i = 0; i <= N; ++i) 66 for (int j = 0; j <= S; ++j) 67 if (f[i][j] < inf) 68 { 69 Min(f[i][j + 1], f[i][j] + D[i]); 70 if (j >= U[i + 1]) 71 Min(f[i + 1][j - U[i + 1]], f[i][j] + j * M); 72 else 73 Min(f[i + 1][0], f[i][j] + (U[i + 1] - j) * D[i + 1] + j * M); 74 } 75 76 int ans = inf; 77 78 for (int i = 0; i <= S; ++i) 79 Min(ans, f[N][i]); 80 81 printf("%d ", ans); 82 } 83
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……
1 #include <cstdio> 2 #include <cstring> 3 4 inline char nextChar(void) 5 { 6 static const int siz = 1024; 7 8 static char buf[siz]; 9 static char *hd = buf + siz; 10 static char *tl = buf + siz; 11 12 if (hd == tl) 13 fread(hd = buf, 1, siz, stdin); 14 15 return *hd++; 16 } 17 18 inline int nextInt(void) 19 { 20 register int ret = 0; 21 register int neg = false; 22 register int bit = nextChar(); 23 24 for (; bit < 48; bit = nextChar()) 25 if (bit == '-')neg ^= true; 26 27 for (; bit > 47; bit = nextChar()) 28 ret = ret * 10 + bit - 48; 29 30 return neg ? -ret : ret; 31 } 32 33 const int siz = 50005; 34 const int inf = 1000000007; 35 36 int tot; 37 int s, t; 38 int hd[siz]; 39 int to[siz]; 40 int fl[siz]; 41 int vl[siz]; 42 int nt[siz]; 43 44 inline void add(int u, int v, int f, int w) 45 { 46 nt[tot] = hd[u]; to[tot] = v; fl[tot] = f; vl[tot] = +w; hd[u] = tot++; 47 nt[tot] = hd[v]; to[tot] = u; fl[tot] = 0; vl[tot] = -w; hd[v] = tot++; 48 } 49 50 int dis[siz]; 51 int pre[siz]; 52 53 inline bool spfa(void) 54 { 55 static int que[siz]; 56 static int inq[siz]; 57 static int head, tail; 58 59 memset(dis, 0x3f, sizeof(dis)); 60 memset(inq, 0, sizeof(inq)); 61 head = 0, tail = 0; 62 que[tail++] = s; 63 pre[s] = -1; 64 dis[s] = 0; 65 inq[s] = 1; 66 67 while (head != tail) 68 { 69 int u = que[head++], v; inq[u] = 0; 70 71 for (int i = hd[u]; ~i; i = nt[i]) 72 if (dis[v = to[i]] > dis[u] + vl[i] && fl[i]) 73 { 74 dis[v] = dis[u] + vl[i], pre[v] = i ^ 1; 75 if (!inq[v])inq[que[tail++] = v] = 1; 76 } 77 } 78 79 return dis[t] < 0x3f3f3f3f; 80 } 81 82 inline int minCost(void) 83 { 84 int cost = 0, flow; 85 86 while (spfa()) 87 { 88 flow = inf; 89 90 for (int i = pre[t]; ~i; i = pre[to[i]]) 91 if (flow > fl[i ^ 1])flow = fl[i ^ 1]; 92 93 for (int i = pre[t]; ~i; i = pre[to[i]]) 94 fl[i] += flow, fl[i ^ 1] -= flow; 95 96 cost += dis[t] * flow; 97 } 98 99 return cost; 100 } 101 102 int n, m, k; 103 104 signed main(void) 105 { 106 n = nextInt(); 107 m = nextInt(); 108 k = nextInt(); 109 110 s = 0, t = n + 1; 111 112 memset(hd, -1, sizeof(hd)); 113 114 for (int i = 1; i <= n; ++i) 115 add(i, t, nextInt(), 0); 116 117 for (int i = 1; i <= n; ++i) 118 add(s, i, inf, nextInt()); 119 120 for (int i = 1; i < n; ++i) 121 add(i, i + 1, k, m); 122 123 printf("%d ", minCost()); 124 }
@Author: YouSiki