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2013
11-11

POJ 2393 Yogurt factory [解题报告] Java

Yogurt factory

问题描述 :

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.

输入:

* 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.

输出:

* 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.

样例输入:

4 5
88 200
89 400
97 300
91 500

样例输出:

126900

温馨提示:

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.

解题代码:

import java.util.Scanner;

public class Main{

public static void main(String[] args) {
   Scanner cin = new Scanner(System.in);
   int n = cin.nextInt();
   int m = n;
   int s = cin.nextInt();
   Node[] node = new Node[n];
   int i = 0;
   long sum = 0;
   while(m > 0) {
    node[i] = new Node(cin.nextInt(), cin.nextInt());
    i++;
    m--;
   }
   for(i=0; i< n-1; i++) {
    if(node[i].c + s < node[i+1].c) {
     node[i+1].c = node[i].c + s;
    }
     sum += node[i].c * node[i].y;
   }
   sum += node[i].c * node[i].y;
   System.out.println(sum);
}

}

class Node {
int c;
int y;
public Node(int c, int y) {
   this.c = c;
   this.y = y;
}
}

  1. 老实说,这种方法就是穷举,复杂度是2^n,之所以能够AC是应为题目的测试数据有问题,要么数据量很小,要么能够得到k == t,否则即使n = 30,也要很久才能得出结果,本人亲测