hdu2639 Bone Collector II--01背包K优解

最新推荐文章于 2020-04-02 16:55:44 发布

LaoJiu_

最新推荐文章于 2020-04-02 16:55:44 发布

阅读量655

点赞数 1

CC 4.0 BY-SA版权

分类专栏：【动态规划】--背包问题

本文链接：https://blog.youkuaiyun.com/LaoJiu_/article/details/51201569

【动态规划】--背包问题专栏收录该内容

11 篇文章

订阅专栏

本文详细介绍了HDU2639问题的解决方法，涉及01背包问题的扩展，求解第K大的骨值最大值。通过构建有序队列合并的思想，实现复杂度优化至O(VNK)，并提供了AC代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=2639

一：原题内容

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 2 ³¹).

Sample Input

Sample Output

二：分析理解
第K优解问题

其基本思想是将每个状态都表示成有序队列，将状态转移方程中的max/min转化成有序队列的合并。这里仍然以01背包为例讲解一下。
首先看01背包求最优解的状态转移方程：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。如果要求第K优解，那么状态f[i][v]就应该是一个大小为K的数组f[i][v][1..K]。其中f[i][v][k]表示前i个物品、背包大小为v时，第k优解的值。 “f[i][v]是一个大小为K的数组”这一句，熟悉C语言的同学可能比较好理解，或者也可以简单地理解为在原来的方程中加了一维。显然f[i][v] [1..K]这K个数是由大到小排列的，所以我们把它认为是一个有序队列。

然后原方程就可以解释为：f[i][v]这个有序队列是由f[i-1][v]和f[i-1][v-c[i]]+w[i]这两个有序队列合并得到的。有序队列 f[i-1][v]即f[i-1][v][1..K]，f[i-1][v-c[i]]+w[i]则理解为在f[i-1][v-c[i]][1..K]的每个数上加上w[i]后得到的有序队列。合并这两个有序队列并将结果的前K项储存到f[i][v][1..K]中的复杂度是O(K)。最后的答案是f[N] [V][K]。总的复杂度是O(VNK)。

01背包再清楚不过了，主要还是是有序队列合并的问题。

三：AC代码

#include<iostream>  
#include<algorithm>  

using namespace std;

int T;
int N, V, K;
int A[35], B[35];
int va[105], vo[105];
int dp[1005][35];

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d%d", &N, &V, &K);

		memset(dp, 0, sizeof(dp));

		for (int i = 1; i <= N; i++)
			scanf("%d", &va[i]);
		for (int i = 1; i <= N; i++)
			scanf("%d", &vo[i]);

		for (int i = 1; i <= N; i++)
		{
			for (int j = V; j >= vo[i]; j--)
			{
				for (int k = 1; k <= K; k++)
				{
					A[k] = dp[j - vo[i]][k] + va[i];
					B[k] = dp[j][k];
				}

				A[K + 1] = -1;
				B[K + 1] = -1;
				int a, b, c;
				a = b = c = 1;

				while (c <= K && (A[a] != -1 || B[b] != -1))
				{
					if (A[a] > B[b])
						dp[j][c] = A[a++];
					else
						dp[j][c] = B[b++];

					if (dp[j][c] != dp[j][c - 1])//去重
						c++;
				}

			}
		}

		printf("%d\n", dp[V][K]);

	}

	return 0;
}