Hdu2639bone Collector II (k-Large backpack)

Bone Collector IITime limit:5000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 3780 Accepted Submission (s): 1947


Problem DescriptionThe Title of this problem be familiar,isn ' t it?yeah,if you had took part in the ' Rookie Cup ' competitio N,you must has seem this title. If you haven ' t seen it before,it doesn ' t matter,i would give you a link:

The link:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today We is not desiring the maximum value of bones,but the k-th maximum value of the bones. NOTICE That,we considerate and ways that get the same value of bones is the same. That means,it'll be a strictly decreasing sequence from the 1st maximum, 2nd maximum. To the k-th maximum.

If the total number of different values are less than k,just ouput 0.
Inputthe first line contain a integer T, the number of cases.
Followed by T cases, each case three lines, the first line contain both integer n, v, K (n <=, v <=, K & lt;=) 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 for each bone.

Outputone integer per line representing the k-th maximum of the total value (this number would be less than 231).

Sample Input

35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1

Sample Output

1220

Authorteddy
Source million Qin Guan end belongs to CHU test instructions on the back of v packaging of the K-major items include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace Std;
int a[10000],b[10000],c[10000],d[10000];
int dp[1000][1000];
int main ()
{
int m,i,j,v1,v2,k;
scanf ("%d", &m);
while (m--)
{
int n;
scanf ("%d%d%d", &n,&v1,&v2);
for (i=0;i<n;i++)
scanf ("%d", &a[i]);
for (i=0;i<n;i++)
scanf ("%d", &b[i]);
Memset (Dp,0,sizeof (DP));
for (i=0;i<n;i++)
{
for (j=v1;j>=b[i];j--)//01 backpack
{
for (k=1;k<=v2;k++)//0~k large backpack to find out.
{
C[k]=dp[j-b[i]][k]+a[i];//c[k] refers to the backpack K large is the item taken.
D[k]=dp[j][k];//d[k] refers to the size of the backpack K large the item is not taken
}
c[k]=d[k]=-1;//done with the end.
int x, y, Z;
X=Y=Z=1;//X is control DP, Y is control C, Z control D.
while (x<=v2&& (c[y]!=-1| | D[Z]!=-1)//start sorting C and D to DP
{
if (C[y]>d[z])
{
Dp[j][x]=c[y];
y++;
}
Else
{
DP[J][X]=D[Z];
z++;
}
if (dp[j][x]!=dp[j][x-1])//Since two K-large does not have an equal
x + +;
}
}
}
printf ("%d\n", Dp[v1][v2]);
}


return 0;
}

Hdu2639bone Collector II (k-Large backpack)