hdu3591The trouble of Xiaoqian 多重背包+完全背包

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//给出Xiaoqian的钱币的价值和其身上有的每种钱的个数//商家的每种钱的个数是无穷,xiaoqian一次最多付20000//问怎样付钱交易中钱币的个数最少//Xiaoqian是多重背包//商家是完全背包#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;    }}

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