POJ-2393 Yogurt factory (贪心）

题目：

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

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思路：

• 这题贪心策略是维护一个值，即每一周的最小成本，这个值等于min（该周的制作成本, 前一周的最小成本 + S），递推并且不需要数组。

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Code:

　　

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

int main() {
    std::ios::sync_with_stdio(false);
    int n, s, cost, amount;
    while(cin>>n>>s) {
        int mincost =  5010;
        long long ans = 0;
        for (int i = 0; i < n; ++i) {
            cin>>cost>>amount;
            mincost = min(mincost+s, cost);
            ans += mincost * amount;
        }
        cout<<ans<<endl;
    }

    return 0;
}