POJ 2393 Yogurt factory

最新推荐文章于 2019-08-08 17:39:21 发布

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本文介绍了一个关于Yucky Yogurt工厂的生产调度问题，通过最优算法确定了每周的酸奶生产和存储策略，以最小化成本并满足客户需求。

Yogurt factory

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week.

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt.

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S.

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

Sample Output

Hint

OUTPUT DETAILS:

In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units.

下面这个代码是别人的，因为过于优美，所以我想学习一下（笑脸）：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("b.txt", "w", stdout);
#define maxn 1000000000
#define N 10010
using namespace std;

int p[N],q[N];
int main()
{
   int n,s;
   long long sum=0;
   while(~scanf("%d%d",&n,&s))
   {
       sum=0;
       for(int i=0;i<n;i++) scanf("%d%d",&p[i],&q[i]);
       for(int i=1;i<n;i++)
       {
           p[i]=min(p[i-1]+s,p[i]);        //这个语句写的太优美了。每周奶酪的最低生产价格是由前一周的价格决定的，因为不管是前多少周总可以计算到前一周，
       }                                //那么如何保证前一周的生产价格是最低的呢？将问题分解，利用记忆化的思想，我们从第一周开始来保证每一周的生产价格都是最低的
       for(int i=0;i<n;i++)
       {
           sum+=p[i]*q[i];
       }
       printf("%lld\n",sum);
   }
   return 0;
}