HDU 2693 Bone Collector II

Problem Description

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 231).

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

记录前k大的最优值，即p[v][k]表示装了v体积的东西获得第k大的价值

多开两个数组a[], b[]来保存 p[j - w[i]][1 -> k] + v[i] ， 以及p[j][1 - >k] 这些值
然后将两个数组里面的值按大到小的顺序重新写入 p[j][1 -k]中

#include <iostream>
#include <string.h>
#define N 110
#define V 1010
using namespace std;
int t,n,v,k,p[V]
,a
,i,j,b
,va
,vo
;
int d;
int x,y,z;
int main()
{
cin>>t;
while(t--)
{
memset(p,0,sizeof(p));
memset(a,-1,sizeof(a));
memset(b,-1,sizeof(b));
cin>>n>>v>>k;
for (i=1; i<=n; i++)
cin>>va[i];
for (i=1; i<=n; i++)
cin>>vo[i];
a[0]=b[0]=1;
for(i=1; i<=n;i++)
{
for(j=v; j>=vo[i]; j--)
{
for (d=1; d<=k; d++)
{
a[d]=p[j-vo[i]][d]+va[i];
b[d]=p[j][d];
}
x=y=z=1;
while(z<=k&&(x<=k||y<=k))
{
if(a[x]>b[y])
p[j][z]=a[x++];
else
p[j][z]=b[y++];
if (p[j][z]!=p[j][z-1])
z++;
}
}
}

cout<<p[v][k]<<endl;
}
return 0;
}