ACM-problem P
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problem:
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. <br> <br>* 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
在产奶时现在的价格和保存费用与未来一周的价格相比,如果花费更少,就扩大生产,把下一周的也产了。
代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
int ci,yi;
}a[10000+10];
int main()
{
int n,s;
int i;
long long sum;
while( cin>>n>>s)
{
for(i=0;i<n;i++)
cin>>a[i].ci>>a[i].yi;
int t=0;
sum=0;
for(i=0;i<n-1;i++)
{
if(t==a[i].yi)
t=0;
else
sum+=a[i].ci*a[i].yi;
if((a[i].ci+s)<a[i+1].ci)
{
sum+=(a[i].ci+s)*a[i+1].yi;
t=a[i+1].yi;
}
}
if(t!=a[n-1].yi)
sum+=a[n-1].ci*a[n-1].yi;
cout<<sum;
}
}
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