HOJ Recoup Traveling Expenses(最长递减子序列变形)

标签：int HOJ Expenses more Recoup bp include dp he

A person wants to travel around some places. The welfare in his company can cover some of the airfare cost. In order to control cost, the company requires that he must submit the plane tickets in time order and the amount of the each submittal must be no more than the previous one. So he must arrange the travel plan according to the airfare cost. The more amount of cost covered with the welfare, the better. If the reimbursement is the same, the more times of flights, the better.

For example, he's route is like this: G -> A-> B -> C -> D -> E -> G, and the quoted price between each destination are as follows:

G -> A: 500
 A -> B: 300
 B -> C: 700
 C -> D: 200
 D -> E: 400
 E -> G: 100

So if he flies from in the order: B -> C, D -> E, E -> G, the reimbursement should be: 

700 + 400 + 100 = 1200 (Yuan)

If the airfare from B to C goes down to 600 Yuan, according to the routine, the reimbursement should be 1100 Yuan. But if he chooses to travel from G -> A, A -> B, C -> D, E -> G, the reimbursement should be: 

500 + 300 + 200 + 100 = 1100 (Yuan)

But in this way, he gets one more flight, so this is a better plan.

Input

The input includes one or more test cases. The first data of each test case is N (1 <= N <= 100), followed by N airfares. Each airfare is integer, between 1 and 2²⁴.

Output

For one test case, output two numbers P and Q. P is the most amount of reimbursement fee. Q is the most times of flights under the circumstances of P.

Sample Input

1 60
2 60 70
3 50 20 70

Sample Output

60 1
70 1
70 2

在求最长递减子序列的基础上变形一下，

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>

using namespace std;
int n;
int dp[105];
int bp[105];
int sp[105];
int a[105];
int ans1,ans2,ans;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);

        memset(dp,0,sizeof(dp));
        memset(bp,0,sizeof(bp));
        memset(sp,0,sizeof(sp));
        ans1=0;ans2=0;ans=0;
        for(int i=1;i<=n;i++)
        {
            int num1=0;
            int num2=0;

            for(int j=i-1;j>=1;j--)
            {
                if(a[i]<=a[j])
                {
                    if(num1<dp[j]||(num1==dp[j]&&num2<bp[j]))
                    {
                        num1=dp[j];
                        num2=bp[j];
                    }
                }
            }
            dp[i]=num1+a[i];
            bp[i]=num2+1;
            if(ans1<dp[i]||(ans1==dp[i]&&ans2<bp[i]))
            {
                ans1=dp[i];
                ans2=bp[i];
            }

        }
        printf("%d %d\n",ans1,ans2);
    }
    return 0;
}

标签：int,HOJ,Expenses,more,Recoup,bp,include,dp,he
From： https://blog.51cto.com/u_15834522/5766299