Bone Collector II（01背包kth）

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: 

Here is the link: http://acm.hdu.edu.cn/showproblem.php?pid=2602 

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum. 

If the total number of different values is less than K,just ouput 0.

Input

The first line contain a integer T , the number of cases. 
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone. 
Output

One integer per line representing the K-th maximum of the total value (this number will be less than 2 ³¹). 
Sample Input

3
5 10 2
1 2 3 4 5
5 4 3 2 1
5 10 12
1 2 3 4 5
5 4 3 2 1
5 10 16
1 2 3 4 5
5 4 3 2 1

Sample Output

12
2
0
题目大意：
输入n,v,k分别代表n个物品，v的体积，以及要求v能装下第k大的价值。
01背包变形，加一维代表第几大，最后dp[v][k]即为答案。

#include <iostream>
#include <cstring>
using namespace std;
int v[105],w[105];
int dp[1005][35],a[35],b[35];
int n,val,k;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(dp,0,sizeof dp);
        cin>>n>>val>>k;
        for(int i=1;i<=n;i++)
            cin>>v[i];
        for(int i=1;i<=n;i++)
            cin>>w[i];
        for(int i=1;i<=n;i++)
            for(int j=val;j>=w[i];j--)
            {
                for(int l=1;l<=k;l++)
                {
                    a[l]=dp[j][l];///不取
                    b[l]=dp[j-w[i]][l]+v[i];//取
                }
                a[k+1]=b[k+1]=-1;
                int x=1,y=1,z=1;
                while(z<=k&&(x<=k||y<=k))///更新，也可以直接排序
                {
                    if(a[x]>b[y])
                        dp[j][z]=a[x++];
                    else
                        dp[j][z]=b[y++];
                    if(dp[j][z]!=dp[j][z-1])
                        z++;
                }
            }
        cout<<dp[val][k]<<'
';
    }
    return 0;
}