[北大机试E]:Yogurt factory(模拟)

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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.

输入

* 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.

输出

* 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.

样例输入

4 5
88 200
89 400
97 300
91 500

样例输出

126900

提示

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.

 模拟题，从最后倒着往前，如果相邻两个week的差大于S，说明在前一周多做然后store是赚的，否则就在这周再做。

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
int n,s;
int c[10005],y[10005];
int main(){
    scanf("%d%d",&n,&s);
    for (int i = 0;i < n;++i)
        scanf("%d%d",c+i,y+i);
    long long ans = 0;
    for (int i = n-1;i > 0;--i){
        if (c[i] - c[i-1] > s){//store
            ans += s*y[i];
            y[i-1] += y[i];
        }else ans += c[i]*y[i];
    } 
    ans += c[0]*y[0];
    printf("%lld
",ans);
    return 0;
}