zoj4019(dp)

Schrödinger's Knapsack

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Time Limit: 1 Second      Memory Limit: 65536 KB

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DreamGrid has a magical knapsack with a size capacity of  called the Schrödinger's knapsack (or S-knapsack for short) and two types of magical items called the Schrödinger's items (or S-items for short). There are  S-items of the first type in total, and they all have a value factor of ; While there are  S-items of the second type in total, and they all have a value factor of . The size of an S-item is given and is certain. For the -th S-item of the first type, we denote its size by ; For the -th S-item of the second type, we denote its size by .

But the value of an S-item remains uncertain until it is put into the S-knapsack (just like Schrödinger's cat whose state is uncertain until one opens the box). Its value is calculated by two factors: its value factor , and the remaining size capacity  of the S-knapsack just after it is put into the S-knapsack. Knowing these two factors, the value of an S-item can be calculated by the formula .

For a normal knapsack problem, the order to put items into the knapsack does not matter, but this is not true for our Schrödinger's knapsack problem. Consider an S-knapsack with a size capacity of 5, an S-item with a value factor of 1 and a size of 2, and another S-item with a value factor of 2 and a size of 1. If we put the first S-item into the S-knapsack first and then put the second S-item, the total value of the S-items in the S-knapsack is ; But if we put the second S-item into the S-knapsack first, the total value will be changed to . The order does matter in this case!

Given the size of DreamGrid's S-knapsack, the value factor of two types of S-items and the size of each S-item, please help DreamGrid determine a proper subset of S-items and a proper order to put these S-items into the S-knapsack, so that the total value of the S-items in the S-knapsack is maximized.

Input

The first line of the input contains an integer  (about 500), indicating the number of test cases. For each test case:

The first line contains three integers ,  and  (), indicating the value factor of the first type of S-items, the value factor of the second type of S-items, and the size capacity of the S-knapsack.

The second line contains two integers  and  (), indicating the number of the first type of S-items, and the number of the second type of S-items.

The next line contains  integers  (), indicating the size of the S-items of the first type.

The next line contains  integers  (), indicating the size of the S-items of the second type.

It's guaranteed that there are at most 10 test cases with their  larger than 100.

Output

For each test case output one line containing one integer, indicating the maximum possible total value of the S-items in the S-knapsack.

Sample Input

3
3 2 7
2 3
4 3
1 3 2
1 2 10
3 4
2 1 2
3 2 3 1
1 2 5
1 1
2
1

Sample Output

23
45
10

Hint

For the first sample test case, you can first choose the 1st S-item of the second type, then choose the 3rd S-item of the second type, and finally choose the 2nd S-item of the first type. The total value is .

For the second sample test case, you can first choose the 4th S-item of the second type, then choose the 2nd S-item of the first type, then choose the 2nd S-item of the second type, then choose the 1st S-item of the second type, and finally choose the 1st S-item of the first type. The total value is .

The third sample test case is explained in the description.

It's easy to prove that no larger total value can be achieved for the sample test cases.

题意：背包问题，但是放的顺序会影响结果。每次放入一物品，其获得的值都可以用v=kr计算，r表示放入后背包剩下的容量，有两种物品，k分别为k1，k2，有大小各不一的这两种物品若干，放入容量为c的背包中，能获得求最大的值。

思路：先放小的优于先放大的，两种物品大小都sort一下，dp[i][j]表示第一种从前j个个中选，第二种从前i个中选后的最大结果。

PS：不要memset这个dp，会tle！

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[2005],b[2005];
ll suma[2005],sumb[2005];
ll dp[2005][2005];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        // memset(dp,0,sizeof(dp));
        memset(suma,0,sizeof(suma));
        memset(sumb,0,sizeof(sumb));
        ll k1,k2,c;
        scanf("%lld%lld%lld",&k1,&k2,&c);
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
        for(int i=1;i<=m;i++)scanf("%lld",&b[i]);
        sort(a+1,a+n+1);
        sort(b+1,b+m+1);
        for(int i=1;i<=n;i++)
        {
            suma[i]=suma[i-1]+a[i];
        }
        for(int i=1;i<=m;i++){
        
            sumb[i]=sumb[i-1]+b[i];
        }
        ll ans=0;
        dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            if(suma[i]<=c)
            {
                dp[0][i]=dp[0][i-1]+k1*(c-suma[i]);
                ans=max(ans,dp[0][i]);
            }
            else dp[0][i]=-1;
        }
        for(int i=1;i<=m;i++)
        {
            if(sumb[i]<=c)
            {
                dp[i][0]=dp[i-1][0]+k2*(c-sumb[i]);
                ans=max(ans,dp[i][0]);
            }
            else dp[i][0]=-1;
        }
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(dp[i-1][j]==-1&&dp[i][j-1]==-1)
                {
                    dp[i][j]=-1;
                }
                if(dp[i-1][j]==-1)
                {
                    dp[i][j]=dp[i][j-1]+k1*(c-sumb[i]-suma[j]);
                }
                else if(dp[i][j-1]==-1)
                {
                    dp[i][j]=dp[i-1][j]+k2*(c-sumb[i]-suma[j]);
                }
                else 
                {
                    dp[i][j]=max(dp[i-1][j]+k2*(c-sumb[i]-suma[j]),dp[i][j-1]+k1*(c-sumb[i]-suma[j]));
                }
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%lld\n",ans);
    }
}