Poj 1276 Cash Machine 多重背包

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.

 

这是一道多重背包问题，题目的意思是有一台取款机，其中有n中不同的面值的金币，dk和nk表示dk这种面值的金币有nk个，求出不超过cash的最大金额的金币大小。应用《背包九讲》的多重背包问题模板就可以求出结果。

 

 #include <iostream>
 #include <cstdio>
 using namespace std;
 #define N 100003
 int c[N],w[N];
 int dp[N];
 int n, v;
 void ZeroOnePack(int cost,int weight)    //01背包
 {
     for(int k=v; k>=cost; k--)
         dp[k] = max(dp[k],dp[k-cost]+weight);
 }
 void CompeletPack(int cost,int weight)    //完全背包
 {
     for(int k=cost; k<=v; k++)
         dp[k] = max(dp[k],dp[k-cost]+weight);
 }
 void MultiplePack(int cost,int weight,int amount)    //多重背包
 {
     if(cost*amount>=v)
     {
         CompeletPack(cost,weight);
         return;
     }
     else
     {
         int k=;
         while(k<amount)
         {
             ZeroOnePack(k*cost,k*weight);
             amount = amount-k;
             k=k*;
         }
         ZeroOnePack(amount*cost,amount*weight);
     }
 }
 int main()
 {
     int i;
     while(scanf("%d %d",&v,&n)!=EOF)
     {
         int totle = ;
         int min = ;
         for(i=; i<n; i++)
         {
             scanf("%d%d",&w[i],&c[i]);    //每种金币的个数和金币面值
             totle += w[i]*c[i];
             if(min>c[i])
                 min = c[i];
         }
         if(totle<v) //如果金币的总量小于输入的v，直接将totle输出
         {
             printf("%d\n",totle);
             continue;
         }
         if(v<min)//如果v小于金币的最小值，则直接输出0
         {
             printf("0\n");
             continue;
         }
         memset(dp,,sizeof(dp));
         for (i=; i<n; i++)
         {
             if(w[i]*c[i] == v)
             {
                 dp[v] = v;
                 break;
             }
             else
                 MultiplePack(c[i],c[i],w[i]);//多重背包
         }

         printf("%d\n",dp[v]);
     }
     return ;
 }