（多重背包+记录路径）Charlie's Change （poj 1787）

http://poj.org/problem?id=1787

 

描述

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. 

输入

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.输出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.".

 

样例输入

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

样例输出

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

 

对于这题说法不一， 有人说是多重背包， 有人说是完全背包， 反正我是当多重背包看了， 但是看看人家写的完全背包还是不难理解的， 挺巧妙的

 

自我感觉第二种解法更方便一些

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;

#define met(a,b) (memset(a,b,sizeof(a)))
#define N 11000
#define INF 0x3f3f3f3f

int pre[N], v[5]={0,1,5,10,25}, dp[N];
int used[N];

/**

dp[i]   代表组成i元时需要的最多的硬币
used[i] 代表第i种硬币用了几次（感觉这点很好，能将多重背包转化为完全背包）
pre[j]  记录的是j从哪个状态转化过来的

*/


int main()
{
    int p, num[5]={0};

    while(scanf("%d%d%d%d%d", &p, &num[1], &num[2], &num[3], &num[4]), p+num[1]+num[2]+num[3]+num[4])
    {
        int i, j, ans[110]={0};

        met(dp, -1);
        met(pre, -1);
        dp[0] = 0;
        for(i=1; i<=4; i++)
        {
            memset(used, 0, sizeof(used));
            for(j=v[i]; j<=p; j++)
            {
                if(dp[j-v[i]]+1>dp[j] && dp[j-v[i]]>=0 && used[j-v[i]]<num[i])
                {
                     dp[j] = dp[j-v[i]]+1;
                     used[j] = used[j-v[i]]+1;
                     pre[j] = j-v[i];
                }
            }
        }

        if(dp[p]<0)
           printf("Charlie cannot buy coffee.\n");
        else
        {
            met(ans, 0);
            i = p;
            while(1)
            {
                if(pre[i]==-1) break;
                ans[i-pre[i]]++;
                i = pre[i];
            }
            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[v[1]], ans[v[2]], ans[v[3]], ans[v[4]]);
        }

    }
    return 0;
}

 

另一种解法：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>

using namespace std;
#define met(a,b) (memset(a,b,sizeof(a)))
const int N=10001;
int dp[N], v[5]={0,1,5,10,25};
int num[N][5];
 //dp[j]为咖啡的价格为j时，所能花费的最多钱币数
 //num[j][i]，表示咖啡的价格为j时，花费的第i种货币的个数
int main()
{
    int a[5], p;

    while(scanf("%d%d%d%d%d", &p, &a[1], &a[2], &a[3], &a[4]), p+a[1]+a[2]+a[3]+a[4])
    {
        int i, j, k;

        met(dp, -1);
        met(num, 0);

        dp[0] = 0;
        for(i=1; i<=4; i++)
        {
            for(j=v[i]; j<=p; j++)
            {
                if(dp[j-v[i]]>=0 && dp[j-v[i]]+1>dp[j] && num[j-v[i]][i]<a[i])
                {
                    dp[j] = dp[j-v[i]] + 1;
                    for(k=1; k<=4; k++)
                    {
                        if(k==i)
                            num[j][k] = num[j-v[i]][k]+1;
                        else
                            num[j][k] = num[j-v[i]][k];
                    }
                }
            }
        }

        if(dp[p]==-1)
            printf("Charlie cannot buy coffee.\n");
        else
            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", num[p][1], num[p][2], num[p][3], num[p][4]);
    }
    return 0;
}