POJ 2393 Yogurt factory

Yogurt factory
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7306 Accepted: 3743
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.

这个做法没错误！暴力找出最小的花费情况，但是没有考虑到时间会超出，这道题原来想的是套用贪心常用的模板，可是居然遇到了贪心常见算法的升级版！开始的时候以为这道题和以前那些题的区别只是不用结构体排序了，每一个的花费总量数据和之前的每一个数据相比，只取最小的那一个，结果却是超时，数据量太大了。所以就有了下面的优化。。。。。

超时代码：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn=10011;int n,s;struct week{    __int64 c;    __int64 y;}week[10005];int main(){   // freopen("S.txt","r",stdin);    while(scanf("%d%d",&n,&s)!=EOF)    {for(int i=1;i<=n;i++)    scanf("%I64d%I64d",&week[i].c,&week[i].y);__int64 sum=0;sum+=week[1].c*week[1].y;    for(int i=2;i<=n;i++)    {        __int64 min=week[i].c*week[i].y;        for(int j=1;j<i;j++)        {            __int64 money=week[i].y*week[j].c+(i-j)*s*week[i].y;            min=(min<money)?min:money;        }        sum+=min;    }printf("%I64d\n",sum);    }    return 0;}

将n的平方优化为 n的效率，将每一周的最小花费都更新，使得每一周的花费都是最小值，而这个最小值就用一个循环将每个周的花费及时更新，看看到底是存几个周核算，还是直接本周生产的单价少。最后再计算花费最小的总量。

#include <iostream>#include <cstdio>#include <math.h>#include <algorithm>using namespace std;struct week{    __int64 c,y;}week[10005];int main(void){  //freopen("S.txt","r",stdin);    int n,s;    scanf("%d%d",&n,&s);        for(int i=1;i<=n;i++)     scanf("%I64d%I64d",&week[i].c,&week[i].y);    __int64 sum=0;    for(int i=2;i<=n;i++)    week[i].c=min(week[i-1].c+s,week[i].c);    for(int i=1;i<=n;i++)        sum+=week[i].c*week[i].y;   printf("%I64d\n",sum);   // printf("%d\n",88*200+89*700+300*5+500*91);    return 0;}