Bone Collector II - HDU 2639 背包第k优解

最新推荐文章于 2021-04-25 16:11:29 发布

提比-我有特殊的AC技巧

最新推荐文章于 2021-04-25 16:11:29 发布

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分类专栏： dp动态规划和递推 HDU

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本文介绍了一道关于01背包问题的编程竞赛题目，旨在寻找第K个最大价值的解决方案。通过逐步构建每种状态的前K个最优解，最终输出所需的第K个最优总价值。

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Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2162 Accepted Submission(s): 1132

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

题意：01背包的第k优解。

思路：每次选出它的前一种状态的前k个和加上它的前k个状态，组成它的前k个状态。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int v[110],w[110],dp[1010][40],n,m,k,f[1010],num;
bool cmp(int a,int b)
{ return a>b; }
int main()
{ int t,i,j,p;
  scanf("%d",&t);
  while(t--)
  { memset(dp,0,sizeof(dp));
    scanf("%d%d%d",&n,&m,&k);
    for(i=1;i<=n;i++)
     scanf("%d",&v[i]);
    for(i=1;i<=n;i++)
     scanf("%d",&w[i]);
    for(i=1;i<=n;i++)
     for(j=m;j>=w[i];j--)
     { for(p=1;p<=k;p++)
       { f[p*2-1]=dp[j][p];
         f[p*2]=dp[j-w[i]][p]+v[i];
       }
       sort(f+1,f+1+2*k,cmp);
       dp[j][1]=f[1];num=1;
       for(p=2;p<=2*k && num<=k;p++)
        if(f[p]!=f[p-1])
         dp[j][++num]=f[p];
     }
    printf("%d\n",dp[m][k]);
  }
}