POJ-1276 Cash Machine(经典多重背包)
2016-06-10 15:24
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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
Sample Output
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.
Source
Southeastern Europe 2002
题意:多重背包,给你一些钱,每一种钱有nk张,要求要你找出不超过cash的最大钱数。
分析:F[i][j]表示前I个物品能不能填到j,然后每个物品枚举一下取多少个,直接暴力DP肯定不可取,会T掉,这种模型有很多种优化方法。
方法1:《挑战程序设计竞赛》中给的方法,改变F[i][j]的含义,此时F[I][J]表示前i种物品得到j时最多能剩下几个i,那么就有
f[i][j] = d[i] (f[i-1][j] >= 0) or f[i][j-w[i]]-1,空间上第一维可以省略掉。
方法2: 《背包九讲》中提到的,多重背包二进制优化法。
每种物品有d[i]个,我们可以把其拆分成log(d[i])种物品(1,2,4,....2^k,d[i] - 2^(k+1)+1),然后做一次01背包。
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.
Source
Southeastern Europe 2002
题意:多重背包,给你一些钱,每一种钱有nk张,要求要你找出不超过cash的最大钱数。
分析:F[i][j]表示前I个物品能不能填到j,然后每个物品枚举一下取多少个,直接暴力DP肯定不可取,会T掉,这种模型有很多种优化方法。
方法1:《挑战程序设计竞赛》中给的方法,改变F[i][j]的含义,此时F[I][J]表示前i种物品得到j时最多能剩下几个i,那么就有
f[i][j] = d[i] (f[i-1][j] >= 0) or f[i][j-w[i]]-1,空间上第一维可以省略掉。
#include <queue> #include <vector> #include <cstdio> #include <utility> #include <cstring> #include <iostream> #include <algorithm> using namespace std; int cash,n,ans,w[1005],d[1005],f[100005]; int main() { cin.sync_with_stdio(false); while(cin>>cash>>n) { memset(f,-1,sizeof(f)); for(int i = 1;i <= n;i++) cin>>d[i]>>w[i]; for(int i = 1;i <= n;i++) { f[0] = d[i]; for(int j = 1;j <= cash;j++) { if(f[j] >= 0) f[j] = d[i]; else if(j >= w[i]) f[j] = f[j-w[i]] - 1; } } ans = 0; for(int j = cash;j;j--) if(f[j] >= 0) { ans = j; break; } cout<<ans<<endl; } }
方法2: 《背包九讲》中提到的,多重背包二进制优化法。
每种物品有d[i]个,我们可以把其拆分成log(d[i])种物品(1,2,4,....2^k,d[i] - 2^(k+1)+1),然后做一次01背包。
#include <queue> #include <vector> #include <cstdio> #include <utility> #include <cstring> #include <iostream> #include <algorithm> using namespace std; int cash,n,ans,w[1005],d[1005]; bool f[100005]; int main() { cin.sync_with_stdio(false); while(cin>>cash>>n) { memset(f,0,sizeof(f)); f[0] = true; for(int i = 1;i <= n;i++) { cin>>d[i]>>w[i]; int k = 1; for(;d[i]-2*k+1 > 0;k*=2) for(int j = cash;j >= k*w[i];j--) f[j] = f[j] || f[j-k*w[i]]; k = d[i]-k+1; for(int j = cash;j >= k*w[i];j--) f[j] = f[j] || f[j-k*w[i]]; } int j = cash; for(;j;j--) if(f[j]) break; cout<<j<<endl; } }
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