POJ3260The Fewest Coins[背包]

The Fewest Coins

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6299		Accepted: 1922

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, C₂ coins 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

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题意：FJ每种硬币有限，售货员无限，最小化交易用的硬币数

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FJ多重背包，售货员完全背包

min一下f[i+m]+d[i]

NOTICE:体积选多大呢？

有一个证明，如果John的付款数大于了maxv*maxv+m，即付硬币的数目大于了maxv，根据鸽笼原理，至少有两个的和对maxv取模的值相等，也就是说，这部分硬币能够用更少的maxv来代替。证毕。

看不懂算了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=,M=*+1e4+,INF=1e9;
int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-; c=getchar();}
    while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
    return x*f;
}
int n,m,om,v[N],c[N],ans=INF,mxv=;
int f[M],d[M];
inline void zp(int v,int w){
    for(int j=m;j>=v;j--) f[j]=min(f[j],f[j-v]+w);
}
inline void cp(int v){
    for(int j=v;j<=m;j++) f[j]=min(f[j],f[j-v]+);
}
inline void mp(int v,int c){
    if(v*c>=m){cp(v);return;}
    int k=;
    while(k<c){
        zp(k*v,k);
        c-=k;
        k*=;
    }
    zp(c*v,c);
}
int main(){
    n=read();m=om=read();
    for(int i=;i<=n;i++) v[i]=read(),mxv=max(mxv,v[i]);m+=mxv*mxv;
    for(int i=;i<=n;i++) c[i]=read();

    for(int i=;i<=m;i++) d[i]=INF;d[]=;
    for(int i=;i<=n;i++)
        for(int j=v[i];j<=m;j++)
            d[j]=min(d[j],d[j-v[i]]+);
    for(int i=;i<=m;i++) f[i]=INF;f[]=;
    for(int i=;i<=n;i++) mp(v[i],c[i]);
    for(int i=;i<=m-om;i++)
        if(f[i+om]+d[i]<ans) ans=f[i+om]+d[i];
    if(ans>=INF) printf("-1");
    else printf("%d",ans);
}