COJ - 1005 - Rent your Airplane and make Money 题解

Description

"ABEAS Corp." is a very small company that owns a single airplane. The customers of ABEAS Corp are large airline companies which rent the airplane to accommodate occasional overcapacity. Customers send renting orders that consist of a time interval and a price that the customer is ready to pay for renting the airplane during the given time period. Orders of all the customers are known in advance. Of course, not all orders can be accommodated and some orders have to be declined. Eugene LAWLER, the Chief Scientific Officer of ABEAS Corp would like to maximise the profit of the company. You are requested to compute an optimal solution.

Small Example: Consider the case where the company has 4 orders: 

• Order 1 (start time 0, duration 5, price 10).
• Order 2 (start time 3, duration 7, price 8).
• Order 3 (start time 5, duration 9, price 7).
• Order 4 (start time 6, duration 9, price 8).

The optimal solution consists in declining Order 2 and 3 and the gain is 10 + 8 = 18. Note that the solution made of Order 1 and 3 is feasible (the airplane is rented with no interruption from time 0 to time 14) but non-optimal.

Input specification

The first line of the input contains a number T (1 <= T <= 30) that indicates the number of test cases to follow. The first line of each test case contains the number of orders n (0 <= n <= 10000). In the following n lines the orders are given. Each order is described by 3 integer values: The start time of the order st (0 <= st < 10^6), the duration d of the order (0 < d < 10^6), and the price p (0 < p < 10^5) the customer is ready to pay for this order.

Output specification

You are required to compute an optimal solution. For each test case your program has to write the total price paid by the airlines.

Sample input

140 5 103 7 145 9 76 9 8

Sample output

18

任务调度问题，不过这里变形了，需要考虑到选择这个任务的最大效益，那么就只能使用动态规划法了，不能使用贪心法了。所以时间效率是O(n^2)。

思想：

1 先按任务开始时间排序

2 然后从后往前使用动态规划法（这里不能从前往后了）

3 如果当前任务结束时间是小于等于后面某个任务，那么就是当前任务可以和后面任务并行执行

4 那么就比较两种情况填表：a 选择当前任务的效益 b 不选择当前任务的效益， 那个大就填那个

这样如下写程序：

#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;struct Task{int st, dur, price;bool operator<(const Task &t) const{return st < t.st;}bool operator<(const int a) const{return st < a;}};//for ( ; j >= 0 && vt[j].st + vt[j].dur > vt[i].st; j--);逻辑错误int tasksPicker(vector<Task> &vt){vector<int> prices(vt.size()+1);prices[vt.size()-1] = vt[vt.size()-1].price;for (int i = vt.size()-2; i >= 0 ; i--){int j=lower_bound(vt.begin()+i+1,vt.end(),vt[i].st+vt[i].dur)-vt.begin();prices[i] = max(prices[i+1], vt[i].price+prices[j]);}return prices[0];}void RentyourAirplaneandmakeMoney(){int T = 0, n = 0, cap;cin>>T;while (T--){cin>>n;vector<Task> vt(n);cap = 0;for (int i = 0; i < n; i++){cin>>vt[i].st>>vt[i].dur>>vt[i].price;cap = max(cap, vt[i].st + vt[i].dur);}sort(vt.begin(), vt.end());cout<<tasksPicker(vt)<<endl;}}int main(){RentyourAirplaneandmakeMoney();return 0;}