hdu2639 Bone Collector II (01背包第k大解）

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

  
  

   
   3
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

  
  

   
   12
2
0
  
  

 

Author

teddy

 

Source

百万秦关终属楚

 

题意：给出一行价值，一行体积，让你在v体积的范围内找出第k大的值；

思路：01背包的进一步优化，普通的背包是求得最大值；

          需要求第k大，我们需要在dp数组上多开一维，记录v体积下第k大的值；

          然后我们利用01背包思想将原值与新得到的值进行按大小合并；

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=105;
int dp[1010][50];
int value[maxn],weight[maxn];
int a[50],b[50];
int n,V,K;
void solve_dp()
{
    memset(dp,0,sizeof(dp));
    for(int i=0;i<n;i++)
        for(int j=V;j>=weight[i];j--)
        {
            for(int k=1;k<=K;k++)
            {
                a[k]=dp[j][k];
                b[k]=dp[j-weight[i]][k]+value[i];
            }
            int x,y,z;
            x=y=z=1;
            a[K+1]=b[K+1]=-1;
            while((z<=K)&&(x<=K||y<=K))  //合并
            {
                if(a[x]>b[y]) dp[j][z]=a[x++];
                else dp[j][z]=b[y++];
                if(dp[j][z]!=dp[j][z-1]) z++;
            }
        }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&V,&K);
        for(int i=0;i<n;i++)
            scanf("%d",&value[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&weight[i]);
        solve_dp();
        printf("%d\n",dp[V][K]);
    }
}