poj_2393 Yogurt factory 贪心

Posted 2021-11-26 dawn-bin

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16669		Accepted: 8176

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. 

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看了几篇博客才有了点思路， 每天生产的酸奶是那一天的，要比较今天的 C 与  Cmin + S  的关系， Cmin是前几个周存储在仓库里的酸奶最小成本， 随着时间的推移，Cin+S ， 直到他不再小于当前的C， 替换掉。

就是一直再权衡当前C与Cmin+S， 求得最小成本。

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#include <iostream>
#include <stdio.h>

using namespace std;


int main()
{
    long long ans;
    int n, s, y, c, minc;
    while(scanf("%d%d", &n, &s)!=-1)
    {
        ans=0; minc=5005;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d", &c, &y);
            if(c > minc +s)
                c = minc + s;
            minc = c;
            ans += c*y;
        }
        printf("%lld\n", ans);
    }
    return 0;
}