01背包（第k优解）

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7165    Accepted Submission(s): 3802

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

 

Author

teddy

 

Source

百万秦关终属楚

 

思路：把放第i个物品看成序列a，不放序列看成序列b，合并依次组成k解。相当于把2的n次方情况排序取前30种。

 

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int dp[1009][35];
int val[1009] , vol[1009];

int main()
{
    int t ;
    scanf("%d" , &t);
    while(t--)
    {
        int n , v , kk , a[35] , b[35];
        scanf("%d%d%d" , &n , &v , &kk);
        memset(dp , 0 , sizeof(dp));
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d" , &val[i]);
        }
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d" , &vol[i]);
        }
        for(int i = 1 ; i <= n ; i++)
        {
            for(int j = v ; j >= vol[i] ; j--)
            {
                for(int k = 1 ; k <= kk ; k++)//将所有解存起来
                {
                    a[k] = dp[j][k];
                    b[k] = dp[j-vol[i]][k] + val[i] ;
                }
                // 对所有解进行排序
                a[kk+1] = -1 ;
                b[kk+1] = -1 ;
                int k = 1 ;
                int x = 1, y = 1 ;
                while(k <= kk && (a[x] != -1 || b[y] != -1))
                {
                    if(a[x] > b[y])
                    {
                        dp[j][k] = a[x];
                        x++ ;
                    }
                    else
                    {
                        dp[j][k] = b[y];
                        y++;
                    }
                    if(dp[j][k] != dp[j][k-1]) k++ ;
                }
            }
        }
        cout << dp[v][kk] << endl ;

    }

    return 0;
}