Cash Machine（多重背包）

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24067		Accepted: 8414

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. 

Notes: 
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc. 

Input

The program input is from standard input. Each data set in the input 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. 

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

Hint

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.

 

题意：以   735 3 4 125 6 5 3 350 这组数据为例，3代表接下来有三组数据，“4 125” 125元的有四张，“6 5” 5元的6张，“3 350” 350元的有3张，问利用这些硬币能凑成的小于735

        并最接近735的面值；

思路：典型的多重背包问题；枚举每一种面值，每次将该面值的使用次数初始化为零，再枚举钱数，当前钱数能达到时，标记已达到，该面值的使用次数加1；

        可是以前做的没印象了，还是理解不深刻，今天又补了补课；

        

 1 #include
 2 #include<string.h>
 3 
 4 int cash,n;
 5 int num[12],p[12],use[1000000];
 6 bool dp[1000100];
 7 
 8 void MultiplePack()
 9 {
10     memset(dp,0,sizeof(dp));
11     dp[0] = 1;//0元肯定能达到，标记为达到；
12     for(int i = 1; i <= n; i++)
13     {
14         memset(use,0,sizeof(use));//第i种面值的使用次数初始化为0；
15         for(int j = p[i]; j <= cash; j++)
16         {
17             if(!dp[j] && dp[j-p[i]] && use[j-p[i]] < num[i])//钱数j能达到，并且面值i的使用次数不超过num[i];
18             {
19                 dp[j] = 1;//标记能达到
20                 use[j] = use[j-p[i]] + 1;//面值i的使用次数加1；
21             }
22         }
23     }
24     for(int i = cash; i >= 0; i--)
25     {
26         if(dp[i])
27         {
28             printf("%d\n",i);
29             break;
30         }
31     }
32 }
33 int main()
34 {
35     while(~scanf("%d",&cash))
36     {
37         scanf("%d",&n);
38         for(int i = 1; i <= n; i++)
39             scanf("%d %d",&num[i],&p[i]);
40         MultiplePack();
41     }
42     return 0;
43 }

View Code

 

        

转载于:https://www.cnblogs.com/LK1994/p/3277286.html