多重背包+记录路径 poj1787题。

题目：

Charlie's Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2459		Accepted: 648

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：有四种面值的钱，已知商品价格，和每种钱的数量，要求选取尽可能多数量的钱，使其总值恰好为商品价格，输出每种面值的钱使用的数量，如果不存在方案，则输出“Charlie cannot buy coffee.”

解题思路：多重背包+打印路径，开始时使用了三重循环，一交，发现tle，后来学习了别人的做法，把每次使用的数量改为数组保存，三重循环变成两重循环，超时代码立即94MS，好神奇。
代码：
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
struct node
{
       int num;
       int id;
       int p;
}pre[15000];
int v[4]={1,5,10,25};
int dp[30000];
int N[1000];
int used[10000];
int main()
{
       int i,j,k,m,n,p,q,C;
       while(~scanf("%d%d%d%d%d",&C,&N[0],&N[1],&N[2],&N[3]))
       {
              if(C==0&&N[0]==0&&N[1]==0&&N[2]==0&&N[3]==0)break;
              for(i=0;i<=C;i++)
              {
                     dp[i]=-1;
                     pre[i].id=i;
                     pre[i].num=pre[i].p=0;
              }
              dp[0]=0;
                 for(i=0;i<4;i++)
                 {
                        memset(used,0,sizeof(used));
                        for(j=v[i];j<=C;j++)
                        if(dp[j-v[i]]+1>dp[j]&&dp[j-v[i]]>=0&&used[j-v[i]]<N[i])
                        {
                               dp[j]=dp[j-v[i]]+1;
                               used[j]=used[j-v[i]]+1;
                               pre[j].id=j-v[i];
                               pre[j].num=1;
                               pre[j].p=v[i];
                        }
                 }
                 if(dp[C]>0)
                 {
                        for(memset(N,0,sizeof(N)),i=C;i;i=pre[i].id)N[pre[i].p]+=pre[i].num;
                        printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",N[1],N[5],N[10],N[25]);
                 }
                 else puts("Charlie cannot buy coffee.");
       }
       return 0;
}