hdu3591The trouble of Xiaoqian 多重揹包+全然揹包

//給出Xiaoqian的錢幣的價值和其身上有的每種錢的個數
//商家的每種錢的個數是無窮,xiaoqian一次最多付20000
//問如何付錢交易中錢幣的個數最少
//Xiaoqian是多重揹包
//商家是全然揹包
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
const int maxn = 20010 ;
const int inf = 0x3f3f3f3f ;
int dp[maxn] ;
int dp_1[maxn] ;
int c[maxn],v[maxn] ;
int n , t ;int cas = 0 ;
int main()
{
    //freopen("in.txt" ,"r" , stdin) ;
    while(scanf("%d%d",&n , &t) &&(n+t))
    {
        for(int i = 1;i <= n;i++)
        scanf("%d" , &v[i]) ;
        for(int i = 1;i <= n;i++)
        scanf("%d" , &c[i]) ;

        for(int i = 1;i < maxn;i++)
        dp[i] = dp_1[i] = inf ;
        dp_1[0] = 0 ; dp[0] = 0 ;

        for(int i = 1;i <= n;i++)
          for(int j = v[i] ; j < maxn ;j++)
          dp_1[j] = min(dp_1[j], dp_1[j-v[i]] + 1) ;

        for(int i = 1;i <= n;i++)
        {
            for(int k = 1;k <= c[i];k*=2)
            {
                for(int j = maxn - 1;j >= k*v[i];j--)
                dp[j] = min(dp[j] , dp[j-k*v[i]] + k) ;
                c[i]-=k;
            }
            for(int j = maxn-1;j >= c[i]*v[i];j--)
            dp[j] = min(dp[j] , dp[j - c[i]*v[i]] + c[i]) ;
        }

        int ans = inf;
        for(int i = t; i <= maxn -10;i++)
        if(dp[i] != inf && dp_1[i - t] != inf)
        ans = min(dp[i] + dp[i-t] , ans) ;

        printf("Case %d: " ,++cas) ;
        if(ans == inf)puts("-1") ;
        else cout<<ans<<endl;
    }
}