hdu Bone Collector II

最新推荐文章于 2021-01-18 14:47:36 发布

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本文探讨了一种特殊的背包问题——求解第K大的骨头价值组合。不同于传统的最大价值求解，该问题需要找到所有可能的价值组合中第K大的价值，并且在实现过程中采用了动态规划的方法来解决这一挑战。

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[sup]31[/sup]).

Sample Input

Sample Output

背包，第k大费用背包问题

#include<iostream>
#include<cmath>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<string.h>
#define INF 2000000;
using namespace std;
int dp[1100][100];
int vis[1000],cs[10000];
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,kt;
        scanf("%d%d%d",&n,&m,&kt);
        for(int i=0;i<n;i++)
            scanf("%d",&vis[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&cs[i]);
        memset(dp,0,sizeof(dp));
        int a[2*kt+10];
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++)
        {
            for(int j=m;j>=cs[i];j--)
            {
                for(int k=0;k<kt;k++)
                {
                    a[2*k]=dp[j-cs[i]][k]+vis[i];//背包大小为j的 装或者不装都保存下来
                    a[2*k+1]=dp[j][k];
                }
                sort(a,a+2*kt,cmp);
                dp[j][0]=a[0];
                int h=1;
                for(int k=1;k<kt;k++)//只取前k个
                {
                    while(1)
                    {
                        if(h>=2*kt||a[h]!=a[h-1])
                            break;
                         h++;
                    }
                    if(h>=2*kt||a[h]==0)
                        break;
                    dp[j][k]=a[h];
                    h++;
                }
            }
        }

        printf("%d\n",dp[m][kt-1]);
    }


}