POJ1787：Charlie's Change(记录路径的多重背包)

Description

Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task.

Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly.

Input

Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.

Output

For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".

Sample Input

12 5 3 1 2
16 0 0 0 1
0 0 0 0 0

Sample Output

Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters.
Charlie cannot buy coffee.

题意：给出总价值，然后给出价值为1，5,10,25的价值的钱币的数量，问能够达到需求的总价值需要的金钱数目最多的方案

思路：乍一看是多重背包，事实上也确实是多重背包，但是由于要输出方案，所以我一直没有想到方法，看了人家的代码，才知道别人记录路径的方法挺神奇的

详细注释在代码中

#include 
#include 
#include 
using namespace std;

const int N = 10010;
int dp[100000],sum,v[5] = {1,5,10,25},ans[30],path[10010];
//dp数组装的是存放钱的总数目，ans则是每种钱的数目，path用来记录路径
void ZeroOnePack(int pos,int k)
{
    int i;
    for(i = sum;i>=k*v[pos];i--)
    {
        if(dp[i-k*v[pos]]+k>dp[i])
        {
            dp[i] = dp[i-k*v[pos]]+k;
            path[i] = pos*N+(i-k*v[pos]);
        }
    }
}

void CompletePack(int pos)
{
    int i;
    for(i = v[pos];i<=sum;i++)
    {
        if(dp[i-v[pos]]+1>dp[i])
        {
            dp[i] = dp[i-v[pos]]+1;
            path[i] = pos*N+(i-v[pos]);
        }
    }
}

void MultiplePack()
{
    int i,j,k;
    for(i = 0;i<4;i++)
    {
        if(v[i]>sum)
        return ;
        if(v[i]*ans[i]>=sum)
        CompletePack(i);
        else
        {
            int cnt = ans[i];
            for(j = 1;j<=cnt;j*=2)
            {
                ZeroOnePack(i,j);
                cnt-=j;
            }
            if(cnt)
            ZeroOnePack(i,cnt);
        }
    }
}

int main()
{
    int i,j,k;
    while(~scanf("%d",&sum),sum)
    {
        for(i = 0;i<4; i++)
            scanf("%d",&ans[i]);
        for(i = 1; i<=sum; i++)
            dp[i] = -N;
        dp[0] = 0;
        MultiplePack();
        if(dp[sum]<0)
        {
            printf("Charlie cannot buy coffee.\n");
            continue;
        }
        memset(ans,0,sizeof(ans));
        while(sum)
        {
            /*由01背包中的path[i] = pos*N+(i-k*v[pos]);可知，s1相当于pos
            s2相当于i-k*v[pos]
            而完全背包pos*N+(i-v[pos]);中，s1相当于pos
            s2相当于i-v[pos]
            */
            int s1 = path[sum]/N;//记录现在改放的是多少钱
            int s2 = path[sum]%N;//记录剩余钱数
            ans[v[s1]]+=(sum-s2)/v[s1];
            sum = s2;
        }
        printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",ans[1],ans[5],ans[10],ans[25]);
    }

    return 0;
}