poj 2393 Yogurt factory——贪心

Yogurt factory

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.

题解： 只需要比较当前星期的价格和前一星期的价格加上存储费用，哪个小就采用哪个

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
struct data
{
    ll p,num,id;   // 价格、数量、编号
}x[10010];

int main()
{
    ll n,s;
    scanf("%lld%lld",&n,&s);
    for (ll i = 0; i < n; i++)
    {
        scanf("%lld%lld",&x[i].p,&x[i].num);
        x[i].id = i;
    }

    ll ans = x[0].p*x[0].num, kase = 0;
    for (ll i = 1; i < n; i++)
    {
        ll tp = x[kase].p+s*(x[i].id-x[kase].id), cur = x[i].p;//一个是之前生产的实际价格，一个是当天生产的价格

        if(cur > tp)
            ans += tp*x[i].num;
        else
        {
            ans += x[i].p*x[i].num;
            kase = i;
        }
    }
    printf("%lld\n",ans);
    return 0;    
}