poj 2393 Yogurt factory(贪心)
2017-07-08 10:49
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Yogurt factory
DescriptionThe 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.
代码:
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long LL; LL dp[10010],num[10010]; int main() { LL n,s; scanf("%I64d%I64d",&n,&s); for(LL i=0; i<n; ++i) scanf("%I64d%I64d",&dp[i],&num[i]); for(LL i=1;i<n;++i) dp[i]=min(dp[i-1]+s,dp[i]); LL ans=0; for(LL i=0; i<n; ++i) ans+=dp[i]*num[i]; printf("%I64d\n",ans); return 0; }
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