poj 3260 hdu 3591 多重背包+完全背包

The Fewest Coins

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 2618 Accepted: 770

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, C₂coins 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

3 705 25 505 2 1

Sample Output

3

View Code

#include<stdio.h>#include<string.h>const int M = 1200000;const int inf = 1<<30;int V,dpsell[M],dpbuy[M];int num[M],w[M];int  min(int a,int b){return a<b?a:b;}void zero_one(int w,int v){    for(int j=V;j>=w;j--)        dpbuy[j]=min(dpbuy[j-w]+v,dpbuy[j]);}int main(){    int n,t,i,j,ca=1;    while(scanf("%d%d",&n,&t),(n||t))    {        for(i=0;i<n;i++) scanf("%d",&w[i]);        for(i=0;i<n;i++) scanf("%d",&num[i]);        V=10000;        dpsell[0]=0,dpbuy[0]=0;        for(i=1;i<=V;i++)            dpsell[i]=dpbuy[i]=inf;        for(i=0;i<n;i++)//售货员有无穷数量的货币，所以先进行一次完全背包            for(j=w[i];j<=V;j++)                dpsell[j]=min(dpsell[j],dpsell[j-w[i]]+1);        for(i=0;i<n;i++)//顾客每种货币都有一定的数量限制，进行一次多重背包        {            int k=1,count=num[i];            while(k<count)            {                zero_one(k*w[i],k);                count-=k;                k<<=1;            }            zero_one(count*w[i],count);        }        int minn=inf;        for(i=t;i<=V;i++)//判断        {            if(dpbuy[i]!=inf&&dpsell[i-t]!=inf&&dpbuy[i]+dpsell[i-t]<minn)              minn=dpbuy[i]+dpsell[i-t];        }        if(minn==inf) printf("Case %d: -1\n",ca++);        else printf("Case %d: %d\n",ca++,minn);    }}