POJ-2393

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11234		Accepted: 5710

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. 

题意：
共有n周，每周生产牛奶花费为c，需求量为y，仓库费用为s，求最小花费。

将s加到每周的成本中去，比较本周花费和上一周花费+s，取最小值。

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

const int MAXN=10010;
int c[MAXN],y[MAXN];

int main(){
    ios::sync_with_stdio(false);
    int n,s;
    while(cin>>n>>s&&n&&s){
        for(int i=0;i<n;i++){
            cin>>c[i]>>y[i];
        }
        long long res=0;
        for(int i=1;i<n;i++){
            c[i]=min(c[i],c[i-1]+s);
        }
        for(int i=0;i<n;i++){
            res+=c[i]*y[i];
        }
        cout<<res<<endl;
    }
    return 0;
}