hdu 2639 Bone Collector II

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1334    Accepted Submission(s): 666 

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 0

 

// dp[s][i][k]  表示在用了 s 空间 i 件物品后排在第k位的值是多少 

dp[s][i][1..k] 在 dp[s][i][1..k] 和 dp[s-s[i]][i-1][1..k] 这2k 个数据中找到前 k 名就是了

就像学校想知道一个年级的前十名 ，只要知道每个班的前十 而过程则是只要比较每2个班的前十、就可以知道全校前十 
#include <iostream>
#include <algorithm>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
int dp[][];
int s[],p[];
int main()
{
    int N,V,K;
    int T;
    scanf("%d",&T);
    while(T--){
         int i,j,k;
         scanf("%d %d %d",&N,&V,&K);
         int t1[],t2[];
         for(i=;i<=N;i++)
            scanf("%d",&p[i]);
         for(i=;i<=N;i++)
            scanf("%d",&s[i]);

         memset(dp,,sizeof(dp));

         for(i=;i<=N;i++){

            for(j=V;j>=s[i];j--){

                for(k=;k<=K;k++)
                {
                    t1[k]=dp[j-s[i]][k]+p[i];
                    t2[k]=dp[j][k];
                }
                t1[k]=t2[k]=-;
                int a=,b=;
                for(k=;(a<=K||b<=K)&&k<=K;){
                    if(t1[a]>t2[b])
                        dp[j][k]=t1[a++];
                    else
                        dp[j][k]=t2[b++];
                    if(dp[j][k]!=dp[j][k-])
                        k++;
                }
            }
         }
         printf("%d\n",dp[V][K]);
    }

    return ;
}