HDU 2639 Bone Collector II

最新推荐文章于 2021-01-18 14:47:36 发布

良月澪二

最新推荐文章于 2021-01-18 14:47:36 发布

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分类专栏：背包问题 HDU 背包问题深入讲解文章标签：动态规划背包 HDU NOIP

本文链接：https://blog.youkuaiyun.com/yandaoqiusheng/article/details/84868995

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本文介绍了一种解决经典背包问题中求第K优解的方法，通过使用动态规划和优化技巧，实现了对不同价值骨头组合的第K最大价值求解。提供了完整的代码实现，并附带了样例输入输出，适合对背包问题和动态规划感兴趣的读者。

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题目链接：传送门
题面：
$B o n e C o l l e c t o r I I$

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;
    }
}