POJ1276Cash Machine[多重背包可行性]

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32971		Accepted: 11950

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

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

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多重背包可行性问题

法1：O(NVlogC)

二进制拆分，w无意义，c和v不开数组也可以

//  poj1276
#include<iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=15,V=1e5+5;
int n,cash,f[V],c[N],w[N],v[N];

inline void zp(int v,int w){
    for(int i=cash;i>=v;i--)
        //f[i]=max(f[i],f[i-v]+w);
        f[i]=f[i]|f[i-v];
}
inline void cp(int v,int w){
    for(int i=v;i<=cash;i++)
//        f[i]=max(f[i],f[i-v]+w);
        f[i]=f[i]|f[i-v];
}
inline void mp(int v,int w,int c){
    if(c*v>=cash){cp(v,w);return;}
    int k=1;
    while(k<c){
        zp(k*v,k*c);
        c-=k;
        k*=2;
    }
    zp(v*c,w*c);
}
 
int main(int argc, const char * argv[]) {
    while(cin>>cash>>n){
        f[0]=1;
        memset(f,0,sizeof(f));    f[0]=1;
        for(int i=1;i<=n;i++){
            cin>>c[i]>>v[i];
            mp(v[i],w[i],c[i]);
        }
        for(int i=cash;i>=0;i--)if(f[i]>0){cout<<i<<"
";break;}
    }
    return 0;
}

法2：O(NV) 然而却比法1慢...............

f[i][j]表示前i种物品装满容量为j的背包后还剩下几个i，不可行为-1

转移先通过f[i-1][j]判断可不可行，再用可行的f[i][j]更新f[i][j+vi](其实就是一遍完全背包)

#include<iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=15,V=1e5+5;
int n,cash,f[N][V],c[N],v[N];

void mpAble(){
    memset(f,-1,sizeof(f));
    f[0][0]=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=cash;j++){
            if(f[i-1][j]>=0) f[i][j]=c[i];
            else f[i][j]=-1;
        }
        for(int j=0;j<=cash-v[i];j++)
            if(f[i][j]>=0)
                f[i][j+v[i]]=max(f[i][j+v[i]],f[i][j]-1);
    }
}
int main(int argc, const char * argv[]) {
    while(cin>>cash>>n){
        for(int i=1;i<=n;i++) cin>>c[i]>>v[i];
        mpAble();
        for(int i=cash;i>=0;i--)if(f[n][i]>=0){cout<<i<<"
";break;}
    }
    return 0;
}