poj 2393 Yogurt factory

http://poj.org/problem?id=2393

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7341		Accepted: 3757

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. 

Source

USACO 2005 March Gold

 

 

 

分析：

简单DP~求最少的花费~

题目是说你每周能够生产牛奶，每周生产的价格为Ci，每周须要上交的牛奶量Yi，你能够选择本周生产牛奶，也可选择提早几周生产出存储在仓库中（仓库无限大，并且保质期不考虑），每一周存仓库牛奶须要花费S元，让你求出全部周的需求量上交的最少花费。

将S转换到花费中~

 

 

AC代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 const int maxn=10011;
 6 int n,s;
 7 int y[maxn],c[maxn];
 8 int main()
 9 {
10     while(scanf("%d%d",&n,&s)!=EOF)
11     {
12     for(int i=0;i<n;i++)
13         scanf("%d%d",&c[i],&y[i]);
14     long long ans=0;
15     for(int i=1;i<n;i++)
16         c[i]=min(c[i-1]+s,c[i]);
17     for(int i=0;i<n;i++)
18         ans+=c[i]*y[i];
19     printf("%lld\n",ans);
20     }
21     return 0;
22 }

 

 

AC代码：

 1 #include<cstdio>
 2 
 3 using namespace std;
 4 int a[10005],b[10005];
 5 
 6 int main() {
 7     int n,s;
 8     while(~scanf("%d %d",&n,&s)) {
 9         for(int i = 0;i < n;i++) {
10             scanf("%d %d",&a[i],&b[i]);
11         }
12         
13         __int64 res = a[0] * b[0];
14         for(int i = 1;i < n;i++) {
15             int mi = 10000000,pos = -1;
16             for(int j = 0;j < i;j++) {
17                 if(mi > s * (i - j) && (a[i] - a[j] > s * (i - j)) ) {
18                     mi = s * (i - j);
19                     pos = j;
20                 }
21             }
22             if(pos != -1)
23                 res += a[pos] * b[i] + mi * b[i];
24             else
25                 res += a[i] * b[i];
26         }
27         printf("%I64d\n",res);
28     }
29     return 0;
30 }

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