Yogurt factory
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 14403 | Accepted: 7213 |
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
思路:
第i个星期的酸奶,可以是第i周生产的,也可以是从i-1周仓库中拿的(i-1周仓库里的奶又可以是第i-1或i-2……第1周生产的)
刚开始用了两个for循环,模拟了每一周的情况,然后就超时了……
学习了一下别人的代码,可以用一个for循环计算每周要运送的酸奶需要的最少的钱,s也算进去=min(等于在该周生产,从上周的仓库里拿)
即:c[i]=min(c[i],c[i-1]+s);
然后再用一个for循环计算答案。
代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,s;
long long c[10005],y[10005];
long long fee[10005];
int main(){
scanf("%d%d",&n,&s);
for(int i=1;i<=n;i++){
scanf("%d%d",&c[i],&y[i]);
if(i!=1){
c[i]=min(c[i-1]+s,c[i]);
}
}
long long ans=c[1]*y[1];
for(int i=2;i<=n;i++){
ans+=c[i]*y[i];
}
printf("%lld\n",ans);
}
超时代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,s;
long long c[10005],y[10005];
long long fee[10005];
int main(){
scanf("%d%d",&n,&s);
for(int i=1;i<=n;i++){
scanf("%d%d",&c[i],&y[i]);
}
fee[1]=c[1]*y[1];
long long ans=fee[1];
for(int i=2;i<=n;i++){
fee[i]=c[i]*y[i];
for(int j=1;j<i;j++){
fee[i]=min(fee[i],(i-j)*s*y[i]+c[j]*y[i]);
}
ans+=fee[i];
}
printf("%lld\n",ans);
}