Yogurt factory
Time Limit: 1000MS |
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Memory Limit: 65536K |
Total Submissions: 9466 |
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Accepted: 4819 |
Descriptionui
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.
Inputspa
* 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.
Outputcode
* 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 Inputorm
4 5
88 200
89 400
97 300
91 500
Sample Outputblog
126900
Hintip
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.
Sourceget
對於第 i 周(i>1),知足酸奶供應,這條最低的單位成本(包括生成成本和存儲成本)爲:min(本週的單位成本Ci , 前一週的最低單位成本+存儲費用S)。
1 #include <cstdio>
2 #include <cstring>
3 using namespace std;
4 const int maxn=10005;
5 int c[maxn],rc[maxn];
6 long long y[maxn],sum;
7 int min(const int &x,const int &y){return x<y?x:y;}
8 int main()
9 {
10 int n,s,cost;
11 while(scanf("%d%d",&n,&s)==2)
12 {
13 sum=0;
14 for(int i=0;i<n;i++)
15 scanf("%d%d",&c[i],&y[i]);
16 memcpy(rc,c,sizeof(rc));
17 rc[0]=c[0];
18 for(int i=0;i<n;i++){
19 cost=rc[i]=min(rc[i],c[i]);
20 for(int j=i+1;j<n;j++){
21 cost+=s;
22 if(cost<c[j]) rc[j]=cost;
23 else break;
24 }
25 }
26 for(int i=0;i<n;i++)
27 sum+=y[i]*rc[i];
28 printf("%lld\n",sum);
29 }
30 }