Bone Collector II（HDU-2639）

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#acm #背包

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本文介绍了一个基于01背包问题的算法实现，旨在寻找物品集合中第K大的价值组合。通过详细的代码示例，展示了如何利用动态规划解决这类问题，并给出了输入输出示例以便理解。

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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：跟之前那个一样，多了个第k大；

思路：01背包+第k最优解；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n,m,k,t;
int dp[1005][35],v[105],cost[105],A[35],B[35];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1; i<=n; i++)
            scanf("%d",&v[i]);
        for(int i=1; i<=n; i++)
            scanf("%d",&cost[i]);
        memset(dp,0,sizeof(dp));
        int a,b,c;
        for(int i=1; i<=n; i++)
        {
            for(int j=m; j>=cost[i]; j--)
            {
                for(int z=1; z<=k; z++)
                {
                    A[z]=dp[j-cost[i]][z]+v[i];  //提取这个包的前k名；
                    B[z]=dp[j][z];  //提取前几个包的前k名；
                }

                A[k+1]=B[k+1]=-1;
                a=b=c=1;
                while(c<=k && (A[a]!=-1 || B[b]!=-1))  //把这个包的前k名跟之前的前k名比较，留下比较之后的前k名；
                {
                    if(A[a]>B[b])
                    {
                        dp[j][c]=A[a];
                        a++;
                    }
                    else
                    {
                        dp[j][c]=B[b];
                        b++;
                    }
                    if(dp[j][c]!=dp[j][c-1]) c++;
                }
            }
        }
        cout<<dp[m][k]<<endl;
    }

    return 0;
}