saketkumarsingh's blog

By saketkumarsingh, 11 months ago, In English

I was trying to solve this question Problem using the concept that answer would be the longest non increasing sequence but it fails on some test cases idk why? Please help me sort it out. Here is my Soln: ~~~~~

#include <bits/stdc++.h>
using namespace std;

int find_lis(vector<int> a) {
    vector<int> dp;
    for (int i : a) {
        int pos = upper_bound(dp.begin(), dp.end(), i)- ``dp.begin();
        if (pos == dp.size()) {
            // we can have a new, longer increasing subsequence!
            dp.push_back(i);
        } else {
            // oh ok, at least we can make the ending element smaller
            dp[pos] = i;
        }
    }
    return dp.size();
}
int main(){
    ifstream fin("cowjog.in");
    ofstream fout("cowjog.out");
    int  n,t;fin>>n>>t;
    vector<vector<int>> v(n, vector<int>(2));
    for(int i=0;i<n;i++) { int x,y;fin>>x>>y; v[i][0]=x, v[i][1]=y; }

    sort(v.begin(), v.end());
    
    vector<int> vec(n);
    for(int i=0;i<n;i++){
        vec[i]=(v[i][0]+t*v[i][1]);
    }
    reverse(vec.begin(), vec.end());
    fout<<find_lis(vec)<<endl;
}

~~~~~

Tags dp, lis
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11 months ago, # |
  Vote: I like it +1 Vote: I do not like it

i think it should be int pos = lower_bound(dp.begin(), dp.end(), i)- ``dp.begin();

  • »
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    11 months ago, # ^ |
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    It isn't, you should have tried it :)

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11 months ago, # |
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LIS isn't the way to go. Here is the answer if you're stuck.

  • »
    »
    11 months ago, # ^ |
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    My bad, LIS actually works

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      »
      11 months ago, # ^ |
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      (Longest Decreasing Sequence over v[i][0]+t*v[i][1])

»
11 months ago, # |
  Vote: I like it -6 Vote: I do not like it

I don't understand dp, but I can only wish you good luck.