hdu 2639求背包第K大值

求第K大值得模板题，只需要多加一维来记录第X大。

然后状态转移由以前的一个方程，变成两个队列的合并，注意合并的时候去重就行

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
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1005;

int N, V, K;
int csti[110],vi[110];
int f[maxn][31];

inline void zoPack(int cost, int weight){
	for(int i=V;i>=cost;--i)
	{	
		int foobar[2*K],ctr = 0;;
		for(int j = 1;j<=K;++j)
		{
			foobar[ctr++] = f[i][j];
			foobar[ctr++] = f[i-cost][j] + weight;
		}
		sort(foobar,foobar + ctr);/*最好这样合并队列*/
		f[i][1] = foobar[--ctr];
		for(int j = 2; j<=K ; ++j)
		{
			while(f[i][j-1] == foobar[--ctr] && ctr >=0); /*注意--的使用..*/ 
			if( ctr < 0) break;
			f[i][j] = foobar[ctr];
		}
	}
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d%d",&N,&V,&K);
		memset(f,0,sizeof(f));

		for(int i=0;i<N;++i)
		scanf("%d",vi+i);
		for(int i=0;i<N;++i)
		scanf("%d",csti+i);
		
		for(int i=0;i<N;++i){
			zoPack(csti[i],vi[i]);
		}
		printf("%d\n",f[V][K]);
	}
	return 0;
}