POJ-2393
Yogurt factory
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11234 Accepted: 5710 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 500Sample Output
126900Hint
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.
题意:
共有n周,每周生产牛奶花费为c,需求量为y,仓库费用为s,求最小花费。
将s加到每周的成本中去,比较本周花费和上一周花费+s,取最小值。
AC代码:
1 //#include<bits/stdc++.h> 2 #include<iostream> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 7 const int MAXN=10010; 8 int c[MAXN],y[MAXN]; 9 10 int main(){ 11 ios::sync_with_stdio(false); 12 int n,s; 13 while(cin>>n>>s&&n&&s){ 14 for(int i=0;i<n;i++){ 15 cin>>c[i]>>y[i]; 16 } 17 long long res=0; 18 for(int i=1;i<n;i++){ 19 c[i]=min(c[i],c[i-1]+s); 20 } 21 for(int i=0;i<n;i++){ 22 res+=c[i]*y[i]; 23 } 24 cout<<res<<endl; 25 } 26 return 0; 27 }