POJ 2393 Yogurt Factory(贪心 ？？ dp)

Yogurt factory
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9380 Accepted: 4768

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

每一周都有单位牛奶的生产费用，和需要向顾客送去的牛奶数量。并且，除了每周生产销售的牛奶外，也可以选择多生产牛奶，用作以后几周的销量，不过这期间需要支付保存的单位费用，问：最少花费。

分析

一个决策的问题。每一周的牛奶都可以决策在之前的周就提前生产出来。
无后效性。上一周的决策结果，不影响下一周的决策。
最优子结构。每一周都可以做出相同的决策过程，尽管结果可能不一样。
很明显是个dp的题。然而为何，大多数的博客都是在写贪心呢，为什么直接比较上一周就可以了。
如果你也这样的疑问就，继续看吧。

很明显，状态转移方程为：
dp[i] = min{dp[i], dp[i-1]+s, dp[i-2]+2*s,…….dp[0] + i * s};
其中dp[i] 表示第i天的最小花费, 初始化为第i周生产的花费。s为保存的花费每一周的。
一直比较，选取前几周中，当周的花费 加上保存至今的 保存花费， 最少的总花费为这一周的花费。
是不是很好理解呢？

将方程化简一下：
dp[i] = min{dp[i], min{dp[i-1] , dp[i-2] +s …, dp[0] + (i-1)*s} +s } 相当于后面的部分提出了个 s；
而 dp[i-1] = min{dp[i-1], dp[i-1] +s, …… dp[0] + (i-1)*s} //有最初的状态转移方程递推出来的。

所以呢， dp[i] = min{dp[i], dp[i-1]+s}
这就得到了很多人写的式子了，因为借用了之前的数据结果，所以看起来很是简单。
而这样的思想，其实在完全背包用滚动数组实现的时候就有所体现了。

这里写出了，是为了说明看似贪心的想法下，是dp思想的体现。

AC代码：

/*===============================================================*   Copyright (C) 2016 All rights reserved.*   *   Filename: I.cpp*   Author:   gsh*   Created_time：2016/8/19 星期五 14:51:15*   Problem_tags:*   Problem_keyword:================================================================*/#include <iostream>#include <cstdio>#include <cmath>typedef long long ll;typedef unsigned long long ull;using namespace std;const int maxn = 1e4 + 100;int c[maxn], y[maxn];int main(){#ifdef LOCAL    freopen("I.in","r",stdin);    freopen("I.out","w",stdout);#endif    int n, s;    while(cin >> n >> s)    {        for(int i = 1; i <= n; i++)            cin >> c[i] >> y[i];        c[0] = 999999;        for(int i = 1; i <= n; i++)            c[i] = min(c[i], c[i-1]+s);        ll ans = 0;        for(int i = 0; i <= n; i++)            ans += c[i]*y[i];        cout << ans << endl;    }    return 0;}

Go big or go home!!!