HDOJ 2639

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1562    Accepted Submission(s): 807

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

 

Recommend

teddy

 

挺有启发意义的一道01背包变形题，让你求出第k大的收益值。我用一个a数组去记录他当前状态下的策略，排序去重，找到前k位记下来，最后可推出结果。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <iomanip>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define mid ((l + r) >> 1)
#define mk make_pair
const int MAXN = 1000 + 50;
const int maxw = 100 + 20;
const int MAXNNODE = 1000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 10007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
const D e = 2.718281828459;

int dp[1010][50];
int a[MAXN];
int volume[MAXN];
int value[MAXN];
int cmp(const int a , const int  b)
{
    return a > b;
}
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // Online_Judge
    int n , m , k;
    int T;
    cin >> T;
    while(T--)
    {
        scanf("%d%d%d" , &n , &m,  &k);
        FOR(i , 0 , n)scanf("%d" , &value[i]);
        FOR(i , 0 , n)scanf("%d" , &volume[i]);
        clr(dp , 0);
        FOR(i , 0  ,n)
        {
            REPP(j , m , volume[i])
            {
                int num = 0;
                FOR(t , 0 , k)
                {
                    a[num++] = dp[j][t];
                    a[num++] = dp[j - volume[i]][t] + value[i];
                }
                sort(a , a + 2 * k , cmp);
                num = 1;
                dp[j][0] = a[0];
                for(int t = 1 ; t < k * 2 && num < k ; t++)
                {
                    if(a[t] != a[t - 1])dp[j][num++] = a[t];
                }
            }
        }
        printf("%d\n" , dp[m][k - 1]);
    }
    return 0;
}