hdu 2639 Bone Collector II（01背包第K大价值）

最新推荐文章于 2025-08-23 19:36:46 发布

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最新推荐文章于 2025-08-23 19:36:46 发布

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原文链接：http://www.cnblogs.com/pshw/p/5033088.html

本文详细介绍了01背包问题的一种变种，即求取第k大价值的方法。通过给出实例输入与输出，展示了如何利用动态规划解决这类问题。包括算法思路、代码实现与代码注释，有助于读者理解并实践该类问题。

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Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3355 Accepted Submission(s): 1726

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

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

Author

teddy

Source

百万秦关终属楚

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还是01背包问题，不过我是第一次写这种类型的题，参考了别人的代码和思想，差不多是一道模板第k大价值题，值得一做。

更新的时候有两个数组A、B，A记录选择第i个物品的从大到小排列，B记录不选择的从大到小排列，然后合并AB，选出AB里面前k个最大的。合并到dp中。。。

题意：第一行三个数，第一个数为物体数量，第二个数总体积，第三个数第k优解。注意！接下来第二行是物体的价值，第三行是物体的体积。

求第k大优解的值是多少。

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int T,i,j,n,m,v,k,kk;
 8     int dp[1005][35],val[1005],vol[1005],A[35],B[35];
 9     scanf("%d",&T);
10     while(T--)
11     {
12         scanf("%d%d%d",&n,&v,&k);
13         for(i=0; i<n; i++)
14             scanf("%d",&val[i]);   //价值
15         for(i=0; i<n; i++)
16             scanf("%d",&vol[i]);    //体积
17         memset(dp,0,sizeof(dp));
18         int a,b,c;
19         for(i=0; i<n; i++)
20             for(j=v; j>=vol[i]; j--)
21             {
22                 for(kk=1; kk<=k; kk++)  //从最优的向后循环
23                 {
24                     A[kk]=dp[j-vol[i]][kk]+val[i];  //选中i物体
25                     B[kk]=dp[j][kk];         //不选i物体
26                 }
27                 A[kk]=-1;   //-1标记为结尾
28                 B[kk]=-1;
29                 a=b=c=1;
30                 while(c<=k&&(A[a]!=-1||B[b]!=-1))
31                 {
32                     if(A[a]>B[b])   //大的数排在前面
33                         dp[j][c]=A[a++];
34                     else
35                         dp[j][c]=B[b++];
36                     if(dp[j][c]!=dp[j][c-1]) //如果两数相同，则只记录一次
37                         c++;
38                 }
39             }
40         printf("%d\n",dp[v][k]);
41     }
42     return 0;
43 }