Bone Collector II

Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3042 Accepted Submission(s): 1578

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
背包求第K大的值,DP[i][j][k]表示放i件物品体积为V的时候第K大的值

for(int j=V;j>=v[i];j--)
{ for(int s=1;s<=k;s++) { A[top++]=Dp[j-v[i]][s]+w[i]; A[top++]=Dp[j][s]; }
}

表示将体积为V时,所有的情况,从中选出前K大的值,对于放每件物品所达到的体积都选出前K大的值,一直贪心到将所有的物品都放完,得到的DP[n][v][k]就是所求的.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAX =  1100;
int Dp[MAX][35];
int w[110],V[110];
int A[35];
int B[35];
int main()
{
    int T;
    int n,v,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d",&n,&v,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&V[i]);
        }
        memset(Dp,0,sizeof(Dp));
        for(int i=1;i<=n;i++)//转化为01背包减少时间复杂度
        {
            for(int j=v;j>=V[i];j--)
            {
                int s;
                for( s=1;s<=k;s++)
                {
                    A[s]=Dp[j-V[i]][s]+w[i];
                    B[s]=Dp[j][s];
                }
                A[s]=-1;
                B[s]=-1;
                int a=1,b=1;
                for(s=1;s<=k&&(A[a]!=-1||B[b]!=-1);)//采用归并的方式,也可以用优先队列
                {
                    if(A[a]>B[b])
                    {
                        Dp[j][s]=A[a];
                        a++;
                    }
                    else
                    {
                        Dp[j][s]=B[b];
                        b++;
                    }
                    if(Dp[j][s]!=Dp[j][s-1])
                    {
                        s++;
                    }
                }
            }
        }
        printf("%d\n",Dp[v][k]);
    }
    return 0;
}