poj 1787 记录路径的多重背包

 

 

 

如题：http://poj.org/problem?id=1787

 

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.

 

 

题目给出总钱数，要求用1,5,10,25价值的硬币拼出总钱数，并要求硬币尽量用的多。

 

一开始没注意到硬币的要求，将硬币价值作为价值w进行背包，wa了好久

关于记录路径，只要在背包的同时，记录这一层是由那一层过来的就行，并同时记录上一层要多少个硬币才能变为这一层，至于哪一种，就可以用int t=(n-path[n])/count[n];推出。

还有要注意的是背包策略，f[j]并不是在一定金额中尽量多放硬币，f[j]代表的是满足拼成j金额能放硬币的最多数量f[j]=max(f[j-k*c]+k) ,初始化f数组为-inf。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int num[5];
int n;
int w[5]={0,1,5,10,25};
int f[10005];
int path[10005];
int count[10005];

void ZeroOnePack(int k,int c)
{
 int j;
 for(j=n;j>=k*c;j--)
 {
  if(f[j-k*c]+k>f[j])
  {
   f[j]=f[j-k*c]+k;
   path[j]=j-k*c;
   count[j]=k;
  }
 }
}
void CompletePack(int c)
{
 int j;
 for(j=c;j<=n;j++)
 {
  int tv=f[j];
  if(f[j-c]+1>f[j])
  {
   f[j]=f[j-c]+1;
   path[j]=j-c;
   count[j]=1;
  }
  
  
 }
}

void MultiplePack(int c,int cnt)
{
 if(c*cnt>=n)
 {
  CompletePack(c);
  return;
 }
 int k=1;
 while(k<cnt)
 {
  ZeroOnePack(k,c);
  cnt-=k;
  k*=2;
 }
  ZeroOnePack(cnt,c);
}

int main()
{
 //freopen("C:\\1.txt","r",stdin);
 //freopen("C:\\3.txt","w",stdout);
 while(~scanf("%d%d%d%d%d",&n,&num[1],&num[2],&num[3],&num[4])
  &&(n||num[1]||num[2]||num[3]||num[4]))
 {
 // printf("%d %d %d %d %d\n",n,num[1],num[2],num[3],num[4]);
  memset(f,0,sizeof(f));
  memset(path,0,sizeof(path));
  memset(count,0,sizeof(count));
  int i;
  for(i=1;i<=n;i++)
   f[i]=-10005;
  for(i=1;i<=4;i++)
   MultiplePack(w[i],num[i]);
  if(f[n]<=0)
   printf("Charlie cannot buy coffee.\n");
  else
  {
   int c1=0,c2=0,c3=0,c4=0;
   while(n)
   {
    int t=(n-path[n])/count[n];
    if(t==1)
     c1+=count[n];
    else if(t==5)
     c2+=count[n];
    else if(t==10)
     c3+=count[n];
    else
     c4+=count[n];
    n=path[n];
   }
   printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",c1,c2,c3,c4);
  }
 }
 return 0;
}