POJ 3260 The Fewest Coins

The Fewest Coins

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3959		Accepted: 1174

Description

Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he receives in change is minimized. Help him to determine what this minimum number is.

FJ wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V₁, V₂, ..., V_N (1 ≤ V_i ≤ 120). Farmer John is carrying C₁ coins of value V₁, C₂coins of value V₂, ...., and C_N coins of value V_N (0 ≤ C_i ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner (although Farmer John must be sure to pay in a way that makes it possible to make the correct change).

Input

Line 1: Two space-separated integers: N and T. 
Line 2: N space-separated integers, respectively V₁, V₂, ..., V_N coins (V₁, ...V_N) 
Line 3: N space-separated integers, respectively C₁, C₂, ..., C_N

Output

Line 1: A line containing a single integer, the minimum number of coins involved in a payment and change-making. If it is impossible for Farmer John to pay and receive exact change, output -1.

Sample Input

3 70
5 25 50
5 2 1

Sample Output

3

Hint

Farmer John pays 75 cents using a 50 cents and a 25 cents coin, and receives a 5 cents coin in change, for a total of 3 coins used in the transaction.

Source

USACO 2006 December Gold

顾客凑齐i需要的最少硬币个数 （多重背包）
店主凑齐i需要的最少硬币个数 （完全背包）
最后统计最少需要多少个硬币。。。。。判断不好上界所以数组开大点 

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

const int INF=0x3f3f3f3f;

int V;

int a[111],c[111];

int dps[20010];

int dpf[20010];

void zpack(int* a,int c,int w)

{

    for(int i=V;i>=c;i--)

    {

        a=min(a,a[i-c]+w);

    }

}

void cpack(int* a,int c,int w)

{

    for(int i=c;i<=V;i++)

    {

        a=min(a,a[i-c]+w);

    }

}

void multipack(int* a,int c,int w,int m)

{

    if(c*m>=V)

    {

        cpack(a,c,1);

        return ;

    }

    int k=1;

    while(k<m)

    {

        zpack(a,c*k,k*w);

        m-=k;

        k*=2;

    }

    zpack(a,c*m,m*w);

}

int main()

{

    int N,T;

    cin>>N>>T;

    for(int i=0;i<N;i++)

        cin>>a;

    for(int i=0;i<N;i++)

        cin>>c;

    for(int i=0;i<T+10000;i++)

        dps=dpf=INF;

    dps[0]=dpf[0]=0;

    V=T+10000;

    for(int i=0;i<N;i++)

        cpack(dps,a,1);

    for(int i=0;i<N;i++)

        multipack(dpf,a,1,c);

    int ans=INF;

    for(int i=T;i<=V;i++)

        ans=min(ans,dpf+dps[i-T]);

    if(ans>=INF) cout<<-1<<endl;

    else

    cout<<ans<<endl;

    return 0;

}