poj2393（贪心）

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

4 588 20089 40097 30091 500

Sample Output

126900

/*觉得题目是相当的难懂，题意是酸奶厂造酸奶的成本会随着牛奶的价格和人工的波动波动，他们自己有个策略可以这个周将下个周的奶存起来但是存起来每个单位要收s cents,给你n个连续的周，然后是每个单位酸奶的成本和需求量求一个最小的花费。贪心的策略：每个周有两个选择要么只造这个周的奶，要么造连着n周的奶，只要在每周造奶前判断一下上个周的成本加上存储费和这个周的造奶价每次维护最小的即可*/

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int main(){    int n, c;    while(scanf("%d%d", &n, &c) != EOF){        int minn = 5005;        int v, m;        long long sum = 0;        scanf("%d%d", &m, &v);        minn = m;        sum += (m*v);        n -= 1;        while(n--){            scanf("%d%d", &m, &v);            if(m < (minn + c)){                sum += m * v;                minn = m;            }            else{                minn = minn + c;    //这里别忘记更新有可能这周的奶是上上上上上上上个周造的那么存储费要累加                sum += minn * v;            }        }        printf("%lld\n", sum);    }    return 0;}