poj3260 The Fewest Coins

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, ...., andC_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

Hint

Farmer John pays 75 cents using a 50 cents and a 25 cents coin, and receives a 5 cents coin in change, for a total of 3 coins used in the transaction.

这题思路是一次枚举钱数，从要买物品的价格到最大值（即上界）,这里最大值为24400（查百度的，用鸽笼原理），然后用num1[m]表示买m元买家最少要用的硬币数（可以用多重背包），num2[m]表示找钱m元最少要用的硬币数（可以用完全背包，因为数量无限），然后设一个值ans，用min（ans，num1[i]+num2[i-jiage])求出最小的硬币数。 

#include<stdio.h>
#include<string.h>
#define inf 88888888
int min(int a,int b){
	return a<b?a:b;
}
int num1[25500],num2[25500],v[105],num[105],w[105];
int main()
{
	int n,i,j,jiage,ans,k,sum;
	int m=24450;
	while(scanf("%d%d",&n,&jiage)!=EOF)
	{
		for(i=1;i<=n;i++){
			scanf("%d",&w[i]);
			v[i]=1;
		}
		for(i=1;i<=n;i++){
			scanf("%d",&num[i]);
		}
		for(j=1;j<=m;j++){
			num1[j]=num2[j]=inf;
		}
		num1[0]=num2[0]=0;
		
		for(i=1;i<=n;i++){
			for(j=w[i];j<=m;j++){
				if(num2[j-w[i]]!=inf){
					num2[j]=min(num2[j],num2[j-w[i]]+v[i]);
				}
			}
			
			ans=num[i]*w[i];
			if(ans>=m){
				for(j=w[i];j<=m;j++){
					if(num1[j-w[i]]!=inf){
						num1[j]=min(num1[j],num1[j-w[i]]+v[i]);
					}
				}
			}
			else{
				k=1;sum=0;
				while(sum+k<num[i]){
					sum+=k;
					for(j=m;j>=k*w[i];j--){
						if(num1[j-k*w[i]]!=inf){
							num1[j]=min(num1[j],num1[j-k*w[i]]+k*v[i]);
						}
					}
					k=k*2;
				}
				k=num[i]-sum;
				for(j=m;j>=k*w[i];j--){
						if(num1[j-k*w[i]]!=inf){
							num1[j]=min(num1[j],num1[j-k*w[i]]+k*v[i]);
						}
				}
			}
		}
		ans=inf;
		for(i=jiage;i<=m;i++){
			if(num1[i]==inf || num2[i-jiage]==inf)continue;
			if(ans>num1[i]+num2[i-jiage]){
				ans=num1[i]+num2[i-jiage];
			}
		}
		if(ans!=inf)
		printf("%d
",ans);
		else printf("-1
");
	}
	return 0;
}