poj 3260（多重+完全背包）

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本博客探讨了农民约翰在购买农场用品时如何通过最小化硬币交易数量实现最优支付策略，包括如何计算所需的最少硬币总数，并提供了解决方案及实例分析。

The Fewest Coins

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6030		Accepted: 1807

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 V ₁, V ₂, ..., V_N coins ( V ₁, ... V_N)
Line 3: N space-separated integers, respectively C ₁, C ₂, ..., 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

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

多重背包求解得到数T(此T非题目给的T因为可以找零所以可以大于它)的最小代价完全背包求解找零后的最小代价前面T开小了直接wa 一怒之下直接标到200000

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
const int inf=(1<<30);
int a[200];
int num[200];
int dp[210015];
int n,t;

void ZeroOnePack(int w,int v)
{
    for(int i=200000;i>=w;i--)
        if(dp[i-w]+v<dp[i])
            dp[i]=dp[i-w]+v;
}

void CompletePack(int w,int v)
{
    for(int i=w;i<=200000;i++)
         if(dp[i-w]+v<dp[i])
            dp[i]=dp[i-w]+v;
}

void MultiplePack()
{
    for(int i=1;i<=n;i++)
    {
        if(a[i]*num[i]>200000)
            CompletePack(a[i],1);
        else
        {
            int k=1;
            while(k<num[i])
            {
                ZeroOnePack(a[i]*k,k);
                num[i]-=k;
                k*=2;
            }
            ZeroOnePack(a[i]*num[i],num[i]);
        }
    }
}

int main()
{
    while(~scanf("%d%d",&n,&t))
    {
        for(int i=1;i<=210005;i++)
            dp[i]=inf;
        dp[0]=0;
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=n;i++)
            scanf("%d",num+i);
        MultiplePack();  //可以凑到的数
        for(int i=1;i<=n;i++)  //完全背包找零
            for(int k=200000;k>=1;k--)
                 if(dp[k+a[i]]+1<dp[k])
                        dp[k]=dp[k+a[i]]+1;
        if(dp[t]==inf)
            puts("-1");
        else
            printf("%d\n",dp[t]);
    }
    return 0;
}