Yogurt factory

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8205   Accepted: 4197

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. 
 
题目是说你每周可以生产牛奶,每周生产的价格为Ci,每周需要上交的牛奶量Yi,你可以选择本周生产牛奶,也可选择提前几周生产出存储在仓库中(仓库无限大,而且保质期不考虑),每一周存仓库牛奶需要花费S元,让你求出所有周的需求量上交的最少花费。
贪心,看了discuss发现只需要考虑邻近的一周,如果a[i].c+s<a[i+1].c,那么本周就把下一周的任务也完成掉。
 
 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
int c,y;
}a[+];
int main()
{
int n,s;
int i;
long long sum;
//freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&s)!=EOF)
{
for(i=;i<n;i++)
scanf("%d%d",&a[i].c,&a[i].y);
int sto=;
sum=;
for(i=;i<n-;i++)
{
if(sto==a[i].y)
sto=;
else
sum+=a[i].c*a[i].y;
if((a[i].c+s)<a[i+].c)
{
sum+=(a[i].c+s)*a[i+].y;
sto=a[i+].y;
}
}
if(sto!=a[n-].y)
sum+=a[n-].c*a[n-].y;
printf("%lld\n",sum);
} }

简单DP:

 #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
int c,y;
}a[+];
int main()
{
int n,s;
int i;
long long sum;
freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&s)!=EOF)
{
sum=;
for(i=;i<n;i++)
scanf("%d%d",&a[i].c,&a[i].y);
for(i=;i<n;i++)
a[i].c=min(a[i].c,a[i-].c+s);
for(i=;i<n;i++)
sum+=a[i].c*a[i].y;
printf("%lld\n",sum);
} }

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