hdu 2639 Bone Collector II

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2306    Accepted Submission(s): 1207

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 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
   
   
   
   

 

/*题解：

01背包+第K优解

*/

#include<cstdio>
#include<cstring>
void solve()
{
	int T,i,j,k,n,m,K,x,y,z,v[110],w[110],d1[110],d2[110],dp[1010][33];//用w[]表示价值 
	scanf("%d",&T);
	while(T--)
	{
		memset(d1,0,sizeof(v));
		memset(d2,0,sizeof(w));
		memset(d1,0,sizeof(d1));
		memset(d2,0,sizeof(d2));
		memset(dp,0,sizeof(dp));
		scanf("%d %d %d",&n,&m,&K);
		for(i=1; i<=n; i++)
			scanf("%d",&w[i]);
		for(i=1; i<=n; i++)
			scanf("%d",&v[i]);
		for(i=1; i<=n; i++)
		{
			for(j=m; j>=v[i]; j--)
			{
				for(k=1; k<=K; k++)//找出前k项 
				{
					d1[k]=dp[j][k];
					d2[k]=dp[j-v[i]][k]+w[i];
				}
				d1[k] = d2[k] = -1;//将每一项末尾置-1 
				x=y=z=1;
				while((x<=K||y<=K)&&(z<=K))
				{
						d1[x]>d2[y]?dp[j][z]=d1[x++]:dp[j][z]=d2[y++];
						if(dp[j][z]!=dp[j][z-1])
							z++; 
				}
			}
		}
		printf("%d\n",dp[m][K]); 
	}
}
int main()
{
	solve();
	return 0;
}