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Time Limit: 1000MS |
Memory Limit: 65536K |
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Total Submissions: 8497 |
Accepted: 4346 |
Description
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
* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.
Output
Sample Input
4 5 88 200 89 400 97 300 91 500
Sample Output
126900
Hint
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.
题意：
有一个奶酪工厂，给出这个工厂每天加工每个奶酪需要的价格，以及每天的需求量，另外，奶酪也可以存放在仓库里（竟然放不坏！），给出每个奶酪存放一天需要的价格，问，这些生产任务全部完成，最少的花费是多少
题解：
刚开始做这道题的时候，感觉自己想的太多了，需要对每天都计算是当天生产最划算还是其他天生产最划算，连优先队列都用上了………
后来写着写着发现有更简便的方法了，完全可以在线性时间内完成运算！
贪心策略：
1，第一天的奶酪只能当天生产，这是无法贪心选取的，暂时假设以后的奶酪全在这一天生产
2，第二天的话，就要考虑是前一天生产存放到第二天还是第二天再生产呢？这就需要比较了，
如果当天生产比较省钱，那么就当天生产！而且假设以后的所有奶酪都在这天生产的！为何这样呢？
因为题目给出的奶酪存放的需要的价格完全一样！（细细想想看）
3，哪天生产更廉价，就哪天生产，这就要求对每天进行编号了。这样才能进行定量计算和查找最优解
另外注意，用64位的整数，个人一个马虎，WA 了一次，好在对自己的贪心思路比较自信，改了一下，果断提交，A掉！
/*
http://blog.csdn.net/liuke19950717
*/
#include<stdio.h>
#include<string.h>
using namespace std;
typedef long long ll;
struct node
{
ll p,num,id;//价格，数量，编号
}x[10005];
ll slove(ll n,ll s)
{
ll ans=x[0].p*x[0].num,kase=0;
for(ll i=1;i<n;++i)
{
ll tp=x[kase].p+s*(i-x[kase].id),cur=x[i].p;//一个是之前生产的实际价格，一个是当天生产的价格
if(tp<cur)//如果之前生产划算
{
ans+=tp*x[i].num;
}
else
{
ans+=cur*x[i].num;
kase=i;
}
}
return ans;
}
int main()
{
ll n,s;
while(~scanf("%lld%lld",&n,&s))
{
for(ll i=0;i<n;++i)
{
scanf("%lld%lld",&x[i].p,&x[i].num);
x[i].id=i;
}
printf("%lld\n",slove(n,s));
}
return 0;
}