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
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
