HDU

 HDU - 2639  Bone Collector II

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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

题意： 求第k优方案 T组样例 给你 n,m,k 有n组数 m大的背包 求第k优方案 下两行分别是 v 和 W

以前求01背包问题的时候都是求的最优的，这道题是求第k大的也就是说得把过程中的不是最优的也得保留下来

所以把dp开成二维 第二维求第k优方案的决策

#include<bits/stdc++.h>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int st=100000;
int dp[1001][40];
int w[1009],v[1009];
int a[1009],b[1009];
int main()
{
    int m,n,k;
    int t;
    cin>>t;
    while(t--)
    {    cin>>n>>m>>k;
        for(int i=0;i<n;i++)
        cin>>v[i];
        for(int i=0;i<n;i++)
        cin>>w[i];
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            for(int j=m;j>=w[i];j--)
            {    
                int kk;
                for( kk=1;kk<=k;kk++) //k种方案
                {
                    a[kk]=dp[j][kk];
                    b[kk]=dp[j-w[i]][kk]+v[i];
                    
                }
                a[kk]=b[kk]=-1;
                int q,w,e;
                q=w=e=1;
                while((a[q]!=-1||b[w]!=-1)&&e<=k)  //相当于把a b数组合并sort一下保留前面k大的，也就是dp[j] 前k种最优方案
                {    
                    if(a[q]>b[w]) dp[j][e]=a[q],q++;
                    else dp[j][e]=b[w],w++;
                    if(dp[j][e]!=dp[j][e-1]) e++;
                    
                }
            
                
            }
            
            
            
            
            
        }
        cout<<dp[m][k]<<endl;
        
    }
    
    return 0;
 }