HDU 2063 Investment （完全背包+数据压缩）

àì夳堔傛蜴生んèń 2022-06-12 01:11 113阅读 0赞

John never knew he had a grand-uncle, until he received the notary’s letter. He learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor.  
  
John did not need that much money for the moment. But he realized that it would be a good idea to store this capital in a safe place, and have it grow until he decided to retire. The bank convinced him that a certain kind of bond was interesting for him.  
  
This kind of bond has a fixed value, and gives a fixed amount of yearly interest, payed to the owner at the end of each year. The bond has no fixed term. Bonds are available in different sizes. The larger ones usually give a better interest. Soon John realized that the optimal set of bonds to buy was not trivial to figure out. Moreover, after a few years his capital would have grown, and the schedule had to be re-evaluated.  
  
Assume the following bonds are available:  
Value Annual interest  
4000 400  
3000 250  
  
With a capital of 10000 10 000 one could buy two bonds of 4000 4 000, giving a yearly interest of 800 800. Buying two bonds of 3000 3 000, and one of 4000 4 000 is a better idea, as it gives a yearly interest of 900 900. After two years the capital has grown to 11800 11 800 , and it makes sense to sell a 3000 3 000 one and buy a 4000 4 000 one, so the annual interest grows to 1050 1 050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is 12850 12 850, which allows for three times 4000 4 000, giving a yearly interest of 1200 1 200.  
  
Here is your problem: given an amount to begin with, a number of years, and a set of bonds with their values and interests, find out how big the amount may grow in the given period, using the best schedule for buying and selling bonds.

Input

The first line contains a single positive integer N which is the number of test cases. The test cases follow.  
  
The first line of a test case contains two positive integers: the amount to start with (at most 1000000 1 000 000), and the number of years the capital may grow (at most 40).  
  
The following line contains a single number: the number d (1 <= d <= 10) of available bonds.  
  
The next d lines each contain the description of a bond. The description of a bond consists of two positive integers: the value of the bond, and the yearly interest for that bond. The value of a bond is always a multiple of $1 000. The interest of a bond is never more than 10% of its value.

Output

For each test case, output – on a separate line – the capital at the end of the period, after an optimal schedule of buying and selling.

Sample Input

1
    10000 4
    2
    4000 400
    3000 250

Sample Output

14050

题解:

一开始想用贪心模拟的，结果发现过程太复杂就放弃了，想起完全背包，试了几次居然能做出答案，然后交上去果断爆了内存，没做过数据压缩题的我居然写出了数据压缩。。。ac了好开心

代码：

#include<iostream>
    #include<stdio.h>
    #include<string>
    #include<cstring>
    #include<map>
    #include<queue>
    #include<stack>
    #include<algorithm>
    #include<math.h>
    #include<deque>
    #include<stack>
    using namespace std;
    struct node
    {
        int p;
        int v;
    };
    node a[15];
    int p[100005];
    int main()
    {
        int i,j,test,n,m,k,num,y,tag,q,d,t;
        double s,bones;//数据压缩后利润可能不为1000的倍数，所以要为double型
        scanf("%d",&test);
        while(test--)
        {
            d=0;
            memset(p,0,sizeof(p));
            scanf("%lf%d",&s,&y);
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%d%d",&a[i].p,&a[i].v);
                a[i].p/=1000;//由于是1000的倍数，就行数据压缩
            }
            d=s;
            s/=1000;//数据压缩
            t=s;//由于数组里面的数字要为整数，所以要转化为整数
            for(q=0;q<y;q++)//进行完全背包
            {
                for(i=0;i<n;i++)
               {
                    for(j=a[i].p;j<=s;j++)
                    {
                        p[j]=max(p[j],p[j-a[i].p]+a[i].v);
                    }
               }
               t=s;
               bones=(double)(p[t])/1000;
               d+=p[t];//d储存s的整数形式
               s+=bones;
            }
            printf("%d\n",d);
        }
        return 0;
    }