Bone Collector II（HDU 2639 DP）

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3471    Accepted Submission(s): 1792

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 2³¹).

 

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

记录每种状态的第k解；

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 int dp[][];
 int val[],vol[];
 int n,v,K;
 int a[],b[];
 int main()
 {
     freopen("in.txt","r",stdin);
     int i,j,k;
     int x,y,z;
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d%d",&n,&v,&K);
         for(i=;i<=n;i++)
             scanf("%d",&val[i]);
         for(i=;i<=n;i++)
             scanf("%d",&vol[i]);
         memset(dp,,sizeof(dp));
         for(i=;i<=n;i++)
         {
             for(j=v;j>=vol[i];j--)
             {
                 for(k=;k<=K;k++)        //记录每种状态的k优解
                 {
                     a[k]=dp[j][k];
                     b[k]=dp[j-vol[i]][k]+val[i];
                 }
                 a[k]=b[k]=-;        //存储的内容已经按照从小到大排序好了，然后合并到dp数组中去
                 x=y=z=;
                 while(z<=K&&(a[x]!=-||b[y]!=-))
                 {
                     if(a[x]>b[y])
                     {
                         dp[j][z]=a[x];
                         x++;
                     }
                     else
                     {
                         dp[j][z]=b[y];
                         y++;
                     }
                     if(dp[j][z-]!=dp[j][z])
                         z++;
                 }
             }
         }
         printf("%d\n",dp[v][K]);
     }
     return ;
 }