hdu2639Bone Collector II（第K大背包）

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3780    Accepted Submission(s): 1947

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

题意就要在V的背包装的第K大物品

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[10000],b[10000],c[10000],d[10000];
int dp[1000][1000];
int main()
{
    int m,i,j,v1,v2,k;
    scanf("%d",&m);
    while(m--)
    {
        int n;
        scanf("%d%d%d",&n,&v1,&v2);
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        for(i=0;i<n;i++)
        scanf("%d",&b[i]);
       memset(dp,0,sizeof(dp));
       for(i=0;i<n;i++)
       {
           for(j=v1;j>=b[i];j--)//01背包
           {
               for( k=1;k<=v2;k++)//把0~K大的背包都求出来。
               {
                    c[k]=dp[j-b[i]][k]+a[i];//c[K]指的是背包第K大是该物品取。
                    d[k]=dp[j][k];//d[k]指的是背包第K大该物品给不取
               }
               c[k]=d[k]=-1;//做结束用的
               int x,y,z;
               x=y=z=1;//x是控制dp的，y是控制c的，z控制d的。
               while(x<=v2&&(c[y]!=-1||d[z]!=-1))//开始将c和d排序给dp
               {
                   if(c[y]>d[z])
                   {
                       dp[j][x]=c[y];
                       y++;
                   }
                   else
                   {
                       dp[j][x]=d[z];
                       z++;
                   }
                   if(dp[j][x]!=dp[j][x-1])//因为两个第K大没有相等的
                    x++;
               }
           }
       }
        printf("%d\n",dp[v1][v2]);
    }


    return 0;
}