[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 350633 4 500 30 6 100 1 5 0 1735 00 3 10 100 10 50 10 10

Sample Output

73563000

題目分析：

多重背包問題。下面列一下背包問題的一般解法

用子問題定義狀態：即f[i][v]表示前i件物品恰放入一個容量為v的背包可以獲得的最大價值。則其狀態轉移方程便是：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。可以壓縮空間，f[v]=max{f[v],f[v-c[i]]+w[i]}

　　這個方程非常重要，基本上所有跟背包相關的問題的方程都是由它衍生出來的。所以有必要將它詳細解釋一下：“將前i件物品放入容量為v的背包中”這個子問題，若只考慮第i件物品的策略（放或不放），那麼就可以轉化為一個只牽扯前i-1件物品的問題。如果不放第i件物品，那麼問題就轉化為“前i-1件物品放入容量為v的背包中”，價值為f[i-1][v]；如果放第i件物品，那麼問題就轉化為“前i-1件物品放入剩下的容量為v-c[i]的背包中”，此時能獲得的最大價值就是f
[i-1][v-c[i]]再加上通過放入第i件物品獲得的價值w[i]。

　　注意f[v]有意義若且唯若存在一個前i件物品的子集，其費用總和為v。所以按照這個方程遞推完畢後，最終的答案並不一定是f[N] [V]，而是f[N][0..V]的最大值。如果將狀態的定義中的“恰”字去掉，在轉移方程中就要再加入一項f[v-1]，這樣就可以保證f[N] [V]就是最後的答案。至於為什麼這樣就可以，由你自己來體會了。

#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 100001int record[11][size];//表示取到第i個元素，總重量小於等於j的最大價值int data[2][11]; //0表示個數  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;elsecout<<record[n-1][sum]<<endl;}}