HDU2639 Bone Collector II

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

 

题意：

每组样例给出背包容量，每个物品的价值，体积，和一个值K，求出第K个最优解；

题解：

01背包扩展，01背包是求最优解，判断每个物品放或者不放时得出最优，这个题可以扩展为，前i个物品中的前K个最优解，将放或者不放的两种情况分别保存，然后取前K个解；排序使用归并排序，并且注意去重。

AC代码

#include 
using namespace std;
int t,n,m,k,v[105],w[1005],dp[1005][32],a[32],b[32];
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--){
        memset(dp,0,sizeof dp);
        cin>>n>>m>>k;
        for(int i=1;i<=n;i++)cin>>v[i];
        for(int i=1;i<=n;i++)cin>>w[i];
        int cnt=0;
        for(int i=1;i<=n;i++)
            for(int j=m;j>=w[i];j--)
        {
            for(int kk=1;kk<=k;kk++){
                a[kk]=dp[j-w[i]][kk]+v[i];
                b[kk]=dp[j][kk];
            }
            a[k+1]=b[k+1]=-1;
            int x,y,z;x=y=z=1;
            while(z<=k&&(x<=k||y<=k)){
                if(a[x]>b[y])dp[j][z]=a[x++];
                else dp[j][z]=b[y++];
                if(dp[j][z]!=dp[j][z-1])z++;
            }
        }
        cout<