hdu 2639 Bone Collector II

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.

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.

One integer per line representing the K-th maximum of the total value (this number will be less than 2³¹).

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

12
2
0
求第k大的。开始想的是加一个从大到小找第k大的值，为什么不对
增加一维表示体积为j时第k大的值。用两个数组分别储存加或不加的两种状态，合并的时候用的是归并的思想。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int n,v,k;
int va[105],vo[105],dp[1010][50];
int a[50],b[50];
int main()
{
    int t,i,j,x,y,z,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&v,&k);
        for(i=0;i<n;i++) scanf("%d",&va[i]);
        for(i=0;i<n;i++) scanf("%d",&vo[i]);
        memset(dp,0,sizeof dp);
        for(i=0;i<n;i++)
            for(j=v;j>=vo[i];j--)
            {
                for(r=1;r<=k;r++)
                {
                    a[r]=dp[j-vo[i]][r]+va[i];
                    b[r]=dp[j][r];
                }
                a[r]=-1,b[r]=-1;
                x=y=z=1;
                while(z<=k&&(a[x]!=-1||b[y]!=-1))
                {
                    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++;
                }
            }
            printf("%d\n",dp[v][k]);
    }
    return 0;
}