Poj 1276 Cash Machine 多重揹包

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26172		Accepted: 9238

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.

 

這是一道多重揹包問題，題目的意思是有一臺取款機，其中有n中不一樣的面值的金幣，dk和nk表示dk這種面值的金幣有nk個，求出不超過cash的最大金額的金幣大小。應用《揹包九講》的多重揹包問題模板就能夠求出結果。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 #define N 100003
 5 int c[N],w[N];
 6 int dp[N];
 7 int n, v;
 8 void ZeroOnePack(int cost,int weight)    //01揹包
 9 {
10     for(int k=v; k>=cost; k--)
11         dp[k] = max(dp[k],dp[k-cost]+weight);
12 }
13 void CompeletPack(int cost,int weight)    //徹底揹包
14 {
15     for(int k=cost; k<=v; k++)
16         dp[k] = max(dp[k],dp[k-cost]+weight);
17 }
18 void MultiplePack(int cost,int weight,int amount)    //多重揹包
19 {
20     if(cost*amount>=v)
21     {
22         CompeletPack(cost,weight);
23         return;
24     }
25     else
26     {
27         int k=1;
28         while(k<amount)
29         {
30             ZeroOnePack(k*cost,k*weight);
31             amount = amount-k;
32             k=k*2;
33         }
34         ZeroOnePack(amount*cost,amount*weight);
35     }
36 }
37 int main()
38 {
39     int i;
40     while(scanf("%d %d",&v,&n)!=EOF)
41     {
42         int totle = 0;
43         int min = 100002;
44         for(i=0; i<n; i++)
45         {
46             scanf("%d%d",&w[i],&c[i]);    //每種金幣的個數和金幣面值
47             totle += w[i]*c[i];
48             if(min>c[i])
49                 min = c[i];
50         }
51         if(totle<v) //若是金幣的總量小於輸入的v，直接將totle輸出
52         {
53             printf("%d\n",totle);
54             continue;
55         }
56         if(v<min)//若是v小於金幣的最小值，則直接輸出0
57         {
58             printf("0\n");
59             continue;
60         }
61         memset(dp,0,sizeof(dp));
62         for (i=0; i<n; i++)
63         {
64             if(w[i]*c[i] == v)
65             {
66                 dp[v] = v;
67                 break;
68             }
69             else
70                 MultiplePack(c[i],c[i],w[i]);//多重揹包
71         }
72 
73         printf("%d\n",dp[v]);
74     }
75     return 0;
76 }

View Code