Yogurt factory（POJ 2393 贪心 or DP）

Posted 2020-06-09

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8205		Accepted: 4197

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

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

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. 

 

题目是说你每周可以生产牛奶，每周生产的价格为Ci，每周需要上交的牛奶量Yi，你可以选择本周生产牛奶，也可选择提前几周生产出存储在仓库中（仓库无限大，而且保质期不考虑），每一周存仓库牛奶需要花费S元，让你求出所有周的需求量上交的最少花费。

贪心，看了discuss发现只需要考虑邻近的一周，如果a[i].c+s<a[i+1].c，那么本周就把下一周的任务也完成掉。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 struct node
 7 {
 8     int c,y;
 9 }a[10000+10];
10 int main()
11 {
12     int n,s;
13     int i;
14     long long sum;
15     //freopen("in.txt","r",stdin);
16     while(scanf("%d%d",&n,&s)!=EOF)
17     {
18         for(i=0;i<n;i++)
19             scanf("%d%d",&a[i].c,&a[i].y);
20         int sto=0;
21         sum=0;
22         for(i=0;i<n-1;i++)
23         {
24             if(sto==a[i].y)
25                 sto=0;
26             else
27                 sum+=a[i].c*a[i].y;
28             if((a[i].c+s)<a[i+1].c)
29             {
30                 sum+=(a[i].c+s)*a[i+1].y;
31                 sto=a[i+1].y;
32             }
33         }
34         if(sto!=a[n-1].y)
35             sum+=a[n-1].c*a[n-1].y;
36         printf("%lld\n",sum);
37     }
38 
39 }

 简单DP:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 struct node
 7 {
 8     int c,y;
 9 }a[10000+10];
10 int main()
11 {
12     int n,s;
13     int i;
14     long long sum;
15     freopen("in.txt","r",stdin);
16     while(scanf("%d%d",&n,&s)!=EOF)
17     {
18         sum=0;
19         for(i=0;i<n;i++)
20             scanf("%d%d",&a[i].c,&a[i].y);
21         for(i=1;i<n;i++)
22             a[i].c=min(a[i].c,a[i-1].c+s);
23         for(i=0;i<n;i++)
24             sum+=a[i].c*a[i].y;
25         printf("%lld\n",sum);
26     }
27 
28 }