[sicily online]1077. Cash Machine
Constraints
Time Limit: 10 secs, Memory Limit: 32 MB
Description
A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of 100 each, 4 bills of 50 each, and 5 bills of 10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Input
The program input is from a text file. Each data set in the file stands for a particular transaction and has the format:
cash N n1 D1 n2 D2 ... nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <= N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k = 1, N. White spaces can occur freely between the numbers in the input. The input data are correct. For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
| Sample Input | Sample Output | Comment |
| 735 3 4 125 6 5 3 350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 10 10 |
735 630 0 0 |
735=1* 350+3* 125+2* 5 630=6* 100+1* 30 or 21* 30 No cash delivered No cash delivered |
The first data set designates a transaction where the amount of cash requested is 735. The machine contains 3 bill denominations: 4 bills of 125, 6 bills of 5, and 3 bills of 350. The machine can deliver the exact amount of requested cash.
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is 630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is 0 and, therefore, the machine delivers no cash.
Output
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
Sample Input
735 3 4 125 6 5 3 350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 10 10
Sample Output
735 630 0 0
题目分析：
多重背包问题。下面列一下背包问题的一般解法
用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。可以压缩空间，f[v]=max{f[v],f[v-c[i]]+w[i]}
这个方程非常重要，基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下：“将前i件物品放入容量为v的背包中”这个子问题，若只考虑第i件物品的策略（放或不放），那么就可以转化为一个只牵扯前i-1件物品的问题。如果不放第i件物品，那么问题就转化为“前i-1件物品放入容量为v的背包中”，价值为f[i-1][v]；如果放第i件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”，此时能获得的最大价值就是f [i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]。 注意f[v]有意义当且仅当存在一个前i件物品的子集，其费用总和为v。所以按照这个方程递推完毕后，最终的答案并不一定是f[N] [V]，而是f[N][0..V]的最大值。如果将状态的定义中的“恰”字去掉，在转移方程中就要再加入一项f[v-1]，这样就可以保证f[N] [V]就是最后的答案。至于为什么这样就可以，由你自己来体会了。
#include<iostream>
#include <iomanip>
#include<stdio.h>
#include<cmath>
#include<iomanip>
#include<list>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#include <sstream>
#include <stack>
#include<queue>
#include<string.h>
#include<set>
using namespace std;
#define size 100001
int record[11][size];//表示取到第i个元素，总重量小于等于j的最大价值
int data[2][11]; //0表示个数 1表示重量
int main()
{
int sum;
while(cin>>sum)
{
memset(record,0,sizeof(record));
memset(data,0,sizeof(data));
int n;
cin>>n;
for(int i=0;i<n;i++)
cin>>data[0][i]>>data[1][i];
for(int j=0;j<=sum&&n!=0;j++)
record[0][j]=min((int)(j/data[1][0]),data[0][0])*data[1][0];
for(int i=1;i<n;i++)
{
for(int j=0;j<=sum;j++)
{
int maxTmp=record[i-1][j];
for(int k=1;k<=data[0][i]&&k*data[1][i]<=j;k++)
{
maxTmp=max(maxTmp,record[i-1][j-k*data[1][i]]+k*data[1][i]);
}
record[i][j]=maxTmp;
}//end j
}
if(n==0)
cout<<0<<endl;
else
cout<<record[n-1][sum]<<endl;
}
}