POJ3260The Fewest Coins题解动态规划DP

The Fewest Coins

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1677		Accepted: 444

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 V ₁, V ₂, ..., V_N coins ( V ₁, ... V_N)
Line 3: N space-separated integers, respectively C ₁, C ₂, ..., 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.

Source

USACO 2006 December Gold

顾客付钱是多重背包用二进制压缩成01背包

店主找零是完全背包

付钱范围刚开始写t+max(v[i]),t+sum(v[i])都是WA

改成t+v[n]*v[n]/k    (1<=k<=4)才AC

#include<cstdio> #define min(a,b) ((a)<(b)?(a):(b)) int n,t,v[105],c[105]; int DP3(int V) { int m=1,ans=1<<20,i,j,k,ww[1005],vv[1005],tt[20005]={0},d[20005]={0}; for(i=0;i<=V;i++)d[i]=tt[i]=1<<20; for(i=1;i<=n;i++) { k=1; j=c[i]; while(j>=k) { j-=k; ww[m]=k; vv[m++]=v[i]*k; k<<=1; } if(j) { ww[m]=j; vv[m++]=j*v[i]; } } d[0]=tt[0]=0; for(i=1;i<m;i++) for(j=V;j>=vv[i];j--) tt[j]=min(tt[j],tt[j-vv[i]]+ww[i]); for(i=1;i<=n;i++) for(j=v[i];j<=V-t;j++) d[j]=min(d[j],d[j-v[i]]+1); for(i=t;i<=V;i++) ans=min(ans,tt[i]+d[i-t]); if(ans>=1<<20) ans=-1; return ans; } int main() { int i; scanf("%d%d",&n,&t); for(i=1;i<=n;i++) scanf("%d",v+i); for(i=1;i<=n;i++) scanf("%d",c+i); printf("%d/n",DP3(t+v[n]*v[n]/4)); }