[HDU 2639]Bone Collector II

Description

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

题解

第$K$大背包。

我们容易想到为背包数组再开一维$k$，

由于序列单调，每次转移的时候归并排序就好。

//It is made by Awson on 2017.9.25
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define sqr(x) ((x)*(x))
using namespace std;
const int N = 100;
const int V = 1000;
const int K = 30;

int n, m, k;
int w[N+5], v[N+5];
int f[V+5][K+5];

void work() {
    scanf("%d%d%d", &n , &m, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
    for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; i++)
    for (int j = m; j >= v[i]; j--) {
        int a[V+5], b[V+5];
        for (int p = 1; p <= k; p++)
        a[p] = f[j][p];
        for (int p = 1; p <= k; p++)
        b[p] = f[j-v[i]][p]+w[i];
        a[k+1] = b[k+1] = -1;
        int la = 1, lb = 1, l = 1;
        while (l<=k && (la <= k || lb <= k)) {
        if (a[la] >= b[lb]) f[j][l] = a[la++];
        else f[j][l] = b[lb++];
        if (f[j][l] != f[j][l-1]) l++;
        }
    }
    printf("%d\n", f[m][k]);
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--)
    work();
    return 0;
}