POJ2393（贪心）

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.

贪心题，只要选c[i],c[i - 1] + s 中比较小的那个即可。

#include <iostream>
#include <algorithm>

using namespace std;
const int maxn = 10000 + 5;

int main()
{
    __int64 sum = 0;
    int n, s, c[maxn], y[maxn];
    cin >> n >> s;
    c[0] = maxn;
    for (int i = 1; i <= n; i++) {
        cin >> c[i] >> y[i];
        c[i] = min (c[i], c[i - 1] + s);
        sum += c[i] * y[i];
    }
    cout << sum << endl;
    return 0;
}