poj 1787 背包+记录路径
http://poj.org/problem?id=1787
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 4512 | Accepted: 1425 |
Description
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
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
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.
Source
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define inf 0x3f3f3f3f
int coin[]={,,,,};
int f[];
int book[];
struct date
{
int num,type;
}Q[];
int main()
{
int C[],P,n,m,i,j,k;
while(cin>>P>>C[]>>C[]>>C[]>>C[]){
if(!(P+C[]+C[]+C[]+C[])) break;
memset(f,-inf,sizeof(f));
memset(Q,,sizeof(Q));
memset(book,,sizeof(book));
int W,L,l=;
f[]=;
for(i=;i<=;++i)
{
l=;
for(k=;C[i];C[i]-=k,k*=){
if(C[i]<k) k=C[i];
// cout<<i<<' '<<k<<endl;
W=k*coin[i];
for(j=P;j>=W;--j)
{ if(f[j-W]!=-inf&&f[j]<f[j-W]+k)
{
f[j]=f[j-W]+k;
Q[j].num=k;
Q[j].type=i;
}
}
}
}//puts("dd");
// cout<<f[P]<<endl;
if(f[P]<) puts("Charlie cannot buy coffee.");
else{
j=P;
i=;
while(j){
book[Q[j].type]+=Q[j].num;
j-=coin[Q[j].type]*Q[j].num;
}
printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",book[],book[],book[],book[]);
}
}
return ;
}
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