HDU 2639 Bone Collector II

题目链接：传送门
题面：
$Bone Collector II$

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

题目大意：

$T$ 组数据，每组数据输入 $N$ ， $V$ ， $k$ ，分别为物品的数量，背包的容量，和你要求的第 $k$ 优解，每种物品只有一个，输出装这个背包的第 $k$ 优解

很经典的求第 $k$ 优解的板子，如果有不大懂的去看我另一篇博客点这里

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define A 1000010
#define B 2010
#define ll long long

using namespace std;
int f[1010][50];
int n, V, k, v[A], w[A], a[A], b[A];

int main() {
    int T;
    cin >> T;
    while (T--) {
        cin >> n >> V >> k;
        for (int i = 1; i <= n; i++) cin >> v[i];
        for (int i = 1; i <= n; i++) cin >> w[i];
        memset(f, 0, sizeof f);
        memset(a, 0, sizeof a);
        memset(b, 0, sizeof b);
        for (int i = 1; i <= n; i++) {
          for (int j = V; j >= w[i]; j--) {
              for (int l = 1; l <= k; l++) {
                  a[l] = f[j][l];
                  b[l] = f[j - w[i]][l] + v[i];
                }
                a[k + 1] = -1;
                b[k + 1] = -1;
                int x = 1, y = 1, o = 1;
                while (o != k + 1 and (a[x] != -1 or b[y] != -1)) {
                    if (a[x] > b[y]) f[j][o] = a[x], x++;
                    else f[j][o] = b[y], y++;
                    if (f[j][o] != f[j][o - 1]) o++;
                }
            }
        }
        cout << f[V][k] << endl;
    }
}