poj 1787 背包+记录路径

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

Charlie's Change
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 4512   Accepted: 1425

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.

Source

CTU Open 2003
  给出了四种不同面值的硬币及其数量，问组成P价值所用的最多数量的硬币是多少，并输出这个方案。应该是多重背包把这个，用二进制优化个数之后就不会T了，刚开始卡了很久因为两层循环我写反了，一直没写过背包有点蒙，由于用到了一维数组的优化，所以对于当前的某件物品，当前的价值用到的子问题必须不涉及到这件物品。
  
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 #define inf 0x3f3f3f3f
 6 int coin[5]={0,1,5,10,25};
 7 int f[10005];
 8 int book[5];
 9 struct date
10 {
11     int num,type;
12 }Q[10005];
13 int main()
14 {
15     int C[5],P,n,m,i,j,k;
16     while(cin>>P>>C[1]>>C[2]>>C[3]>>C[4]){
17         if(!(P+C[1]+C[2]+C[3]+C[4])) break;
18         memset(f,-inf,sizeof(f));
19         memset(Q,0,sizeof(Q));
20         memset(book,0,sizeof(book));
21         int W,L,l=0;
22         f[0]=0;
23         for(i=1;i<=4;++i)
24         {
25             l=0;
26             for(k=1;C[i];C[i]-=k,k*=2){
27                 if(C[i]<k) k=C[i];
28                // cout<<i<<' '<<k<<endl;
29                 W=k*coin[i];
30             for(j=P;j>=W;--j)
31             {
32 
33 
34                     if(f[j-W]!=-inf&&f[j]<f[j-W]+k)
35                         {
36                             f[j]=f[j-W]+k;
37                             Q[j].num=k;
38                             Q[j].type=i;
39                         }
40             }
41             }
42         }//puts("dd");
43        // cout<<f[P]<<endl;
44        if(f[P]<0) puts("Charlie cannot buy coffee.");
45        else{
46             j=P;
47             i=0;
48             while(j){
49                 book[Q[j].type]+=Q[j].num;
50                 j-=coin[Q[j].type]*Q[j].num;
51             }
52             printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",book[1],book[2],book[3],book[4]);
53        }
54     }
55     return 0;
56 }

 

posted @ 2017-11-02 22:33  *zzq  阅读(195)  评论(0)  编辑  收藏  举报