POJ-3260 The Fewest Coins

The Fewest Coins

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 6615	Accepted: 2039

Description

Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he receives in change is minimized. Help him to determine what this minimum number is.

FJ wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V₁, V₂, ..., V_N (1 ≤ V_i ≤ 120). Farmer John is carrying C₁ coins of value V₁, C₂ coins of value V₂, ...., and C_N coins of value V_N (0 ≤ C_i ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner (although Farmer John must be sure to pay in a way that makes it possible to make the correct change).

Input

Line 1: Two space-separated integers: N and T.
Line 2: N space-separated integers, respectively V₁, V₂, ..., V_N coins (V₁, ...V_N)
Line 3: N space-separated integers, respectively C₁, C₂, ..., C_N

Output

Line 1: A line containing a single integer, the minimum number of coins involved in a payment and change-making. If it is impossible for Farmer John to pay and receive exact change, output -1.

Sample Input

3 70
5 25 50
5 2 1

Sample Output

3

 1 #include<iostream>
 2 #include<string.h>
 3 #include<stdio.h>
 4 #include<algorithm>
 5 #define INF 999999999
 6 using namespace std;
 7 int dp1[20005],dp2[20005];
 8 int main()
 9 {
10     int n,m;
11     while(cin>>n>>m)
12     {
13         memset(dp1,0,sizeof(dp1));
14         memset(dp2,0,sizeof(dp2));
15         int val[150],num[150];
16         for(int i=1;i<=n;i++)
17             cin>>val[i];
18         for(int i=1;i<=n;i++)
19             cin>>num[i];
20         for(int i=1;i<=20000;i++)
21             dp1[i]=dp2[i]=INF;
22         dp1[0]=dp2[0]=0;
23         for(int i=1;i<=n;i++)
24         {
25             for(int k=1,flag=0;;k*=2)
26             {
27                 if(k*2>num[i])
28                 {
29                     k=num[i]-k+1;
30                     flag=1;
31                 }
32                 for(int j=20000;j>=k*val[i];j--)
33                 {
34                     dp1[j]=min(dp1[j],dp1[j-k*val[i]]+k);
35                 }
36                 if(flag)
37                     break;
38             }
39         }
40         for(int i=1;i<=n;i++)
41         {
42             for(int j=val[i];j<=20000;j++)
43                 dp2[j]=min(dp2[j],dp2[j-val[i]]+1);
44         }
45         int ans=INF;
46         for(int i=m;i<=20000;i++)
47         {
48             ans=min(ans,dp1[i]+dp2[i-m]);
49         }
50         if(ans==INF)
51         cout<<-1<<endl;
52         else
53             cout<<ans<<endl;
54     }
55     return 0;
56 }

 

转载于:https://www.cnblogs.com/ISGuXing/p/7215280.html