Light oj 1277 - Looking for a Subsequence（LIS）

1277 - Looking for a Subsequence

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Given a sequence of integers S = {a₁, a₂, a₃ ... a_n} and an integer m, you have to find a subsequence of S containing m elements, P = {b₁, b₂, b₃ ... b_m} such that (b₁ < b₂ < ... < b_m). If there are several subsequences possible then you should find the one where b₁ is leftmost. If there is still a tie, then check for the one where b₂ is leftmost and so on.

For example, let the sequence be, 8 7 5 6 5 1 2 7 and m = 2. Then (1, 2), (5, 6), (5, 7), (1, 7) ... etc can be solutions. But we are looking for (5, 6).

Input

Input starts with an integer T (≤ 12), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 10) where q denotes the total number of queries. The next line contains n space separated integers denoting a₁, a₂ ... a_nrespectively. Each of the next q lines contains an integer m (1 ≤ m ≤ n). You can assume that -10⁸ ≤ a_i ≤ 10⁸.

Output

For each case, print the case number in a line. Then for each query print a single line containing the desired subsequence or 'Impossible' if there is no such subsequence. Insert a single space between two integers of a subsequence.

Sample Input	Output for Sample Input
3 6 3 3 4 1 2 3 6 6 4 2 6 2 2 4 6 1 3 5 3 4 8 5 8 7 5 6 5 1 2 7 1 2 3 4 5	Case 1: Impossible 1 2 3 6 3 4 Case 2: 2 4 6 Impossible Case 3: 8 5 6 5 6 7 Impossible Impossible

Note

Dataset is huge, use faster I/O methods.

---

PROBLEM SETTER: JANE ALAM JAN

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-12

typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define N 100005

int n,m;
int len;
int dp[N],a[N];
int vis[N];
vector<int>ans;

int seach(int x)
{
    int le,ri;
    le=0;
    ri=len;
    int ans=0;
    while(le<=ri)
    {
       int mid=MID(le,ri);
       if(dp[mid]>x)
       {
           ans=mid;
           le=mid+1;
       }
       else
         ri=mid-1;
    }
   return ans+1;
}


void solve(int x)
{
    int i,j;
    ans.clear();

    for(i=1;i<=n;i++)
        if(vis[i]>=x) break;

    int len=x;
    int pre=-1000000000;

    while(1)
    {
        if(vis[i]>=len&&a[i]>pre)
        {
            ans.push_back(a[i]);
            pre=a[i];
            len--;
        }
        i++;
        if(ans.size()>=x) break;
    }

    for(int i=0;i<ans.size();i++)
    {
        if(i) printf(" ");
        printf("%d",ans[i]);
    }
  printf("\n");
}

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        dp[0]=1000000000;
        len=0;

        for(int i=n;i>=1;i--)
        {
            int tt=seach(a[i]);
            dp[tt]=a[i];
            if(tt>len) len=tt;
            vis[i]=tt;
        }
        printf("Case %d:\n",++ca);
        int x;
        while(m--)
        {
            scanf("%d",&x);
            if(x>len)
                printf("Impossible\n");
            else
                solve(x);
        }
    }
  return 0;
}