POJ 2063 Investment【经典完全背包】

DP背包（注意减少内存的转换）

快速链接：http://poj.org/problem?id=2063

CSUST 2012年暑假8月组队后第八次个人赛：http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=11473#problem/D

Investment

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 5370 Accepted: 1848

Description

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: 
ValueAnnual
interest4000
3000400
250
With a capital of e10 000 one could buy two bonds of $4 000, giving a yearly interest of $800. Buying two bonds of $3 000, and one of $4 000 is a better idea, as it gives a yearly interest of $900. After two years the capital has grown to $11 800, and it makes sense to sell a $3 000 one and buy a $4 000 one, so the annual interest grows to $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 $12 850, which allows for three times $4 000, giving a yearly interest of $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 $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

110000 424000 4003000 250

Sample Output

14050

Source

Northwestern Europe 2004

下面的是以前写的，现在重学DP 又重新写了一遍。

题意：

     首先输入一个整数 T ，表示有 T 组测试数据。

     给你一定本金 V 和存入银行的年限 year ( 也就是样例中第二行输入的数据 )

     然后输入一个整数 n 表示有 n 种年利息计算法。

     剩下的 n 行就是每种本金产生的年利息的钱。

算法思想：完全背包 第二个基本的背包问题模型，每种物品可以放无限多次。PS：如果你不懂什么是完全背包请看 

http://blog.csdn.net/cfreezhan/article/details/7865136#_P02:_完全背包问题

完全背包状态转移方程：

for i=1..N

    for v=0..V

        f[v]=max{f[v],f[v-cost]+weight}

注意：转换 本题的背包容量就是本金

            但是每一年都有利息产生，所以本金都会增长，从而背包的体积不断增大。

            放入背包的物品的容量就是每种利息所需的本金cost[i]，放入物品后产生的价值就是  

           利息weight[i]。

内存问题：注意题目中的本金很大但是最开始最多的本金不超过$1 000 000

                     而且每种计算利息所需的本金都是$1 000的倍数，所以为了避免超内存，每次  

                     dp时先把本金除以$1 000，相应的cost也除以$1 000

那么dp数组到底开多大才合适呢，题中说最多存40年，开始时本金不超过$1 000 000而每次的利息又不会超过本金的 10% ，

所以 数组的大小应该是 $1 000 000 / 1000 * ( 1.1 ) ^40

而 1.1^40 大概为 45.259256，就取 50 吧。

从而数组开 1 000 * 50 = 50 000

//Accepted360 KB79 msC++667 B2013-03-14 19:26:23#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 50 * 1000;int dp[maxn];int main(){int T;int V, year;int n;int cost[11], weight[11];scanf("%d", &T);while(T--){scanf("%d%d", &V, &year);int Max = V;scanf("%d", &n);for(int i = 1; i <= n; i++){scanf("%d%d", &cost[i], &weight[i]);cost[i] /= 1000; //每种债券除以1000减少内存。}while(year--){V /= 1000; //相应的每次的本金除以1000减少内存memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; i++){for(int j = cost[i]; j <= V; j++)dp[j] = max(dp[j], dp[j-cost[i]] + weight[i]);}Max += dp[V]; //本金+利息(注意利息开始没有除以也不用除以1000,所以Max即为所求)V = Max;}printf("%d\n", V);}return 0;}

下面的是以前写的。。。思路完全一样，只是上面重做时重新分析了下数组大小问题。。。

重点:关于内存的处理,比赛时一直错就是内存没处理好。

        后来看了学长的题解(http://www.shabiyuan.com/?category欢迎访问)

        才清楚。具体分析见下面的代码区~~~~~~

//AC 948k 94ms C++ POJ 2063 Investment/*思路:DPz之完全背包(如果不懂完全背包的推荐看我转载的《背包九讲》的博客，推荐一道完全背包入门题目hdu 1114)       完全背包:即每种物品可以放N次。       状态转移方程:if(dp[j]<dp[j-value[i]]+interest[i])dp[j]=dp[j-value[i]]+interest[i];       需要注意的是，这道题目是背包体积不断增大的完全背包。    开始背包的体积是本金，每隔一年都会得到一定的利息，所以每年背包的体积都会变大。       重点:关于内存的处理,比赛时一直错就是内存没处理好,后来看了学长的题解(http://www.shabiyuan.com/?category欢迎访问)才清楚。    由于本金的数值很大，而数组又不能开的特别大。所以用dp处理时,要把每次的本金除以1000减少内存，相应的每种债券所需要的钱也除以1000*/

//400k 63ms (当数组开到maxn=60000时)

#include<cstdio> #include<cstring>const int maxn=200000;int dp[maxn];int main(){int test;int start,year;int d;int i,j;int value[15],interest[15];int ans,max;scanf("%d",&test);while(test--){scanf("%d%d",&start,&year);ans=max=start;scanf("%d",&d);for(i=0;i<d;i++){scanf("%d%d",&value[i],&interest[i]);value[i]/=1000;//每种债券除以1000减少内存。}while(year--){memset(dp,0,sizeof(dp));max/=1000; //相应的每次的本金除以1000减少内存for(i=0;i<d;i++)for(j=value[i];j<=max;j++)if(dp[j]<dp[j-value[i]]+interest[i])dp[j]=dp[j-value[i]]+interest[i];ans+=dp[max];//本金+利息(注意利息开始没有除以也不用除以1000,所以ans即为所求)max=ans;}printf("%d\n",ans);}return 0;}