Back to Underworld Lightoj 1009 | An Easy Bfs/Dfs Problem

Problem link

Lets put a node is vampire, then all the nodes adjacent with it will be lykans.

Thats why we just need to traverse all the node by a simple bfs or dfs, by toggling nodes states( if parents is vampire then it will be lykans)

Code:

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#include<bits/stdc++.h> using namespace std; #define f0(n) for(int i=0;i<n;i++) #define pb push_back #define IOS ios::sync_with_stdio(false); const int maxn=20004; vector<int>adj[maxn]; int vis[maxn]; ///solution void init(){ f0(maxn){ adj[i].clear(); vis[i]=0; } } int bfs(int x){ queue<int>q; int va=0,ly=0; q.push(x); vis[x]=1; va++; while(!q.empty()){ int u=q.front(); q.pop(); for(auto v:adj[u]){ if(vis[v]==0){ if(vis[u]==1){ vis[v]=2; ly++; } else {vis[v]=1;va++;} q.push(v); } } } return max(va,ly); } int tc=0; void solution(){ init(); int n; cin>>n; f0(n){ int u,v; cin>>u>>v; adj[u].pb(v); adj[v].pb(u); } int ans=0; for(int i=1;i<=20000;i++){ if(vis[i]==0 && !adj[i].empty()){ ans+=bfs(i); } } cout << "Case "<<++tc << ": " << ans << endl; } signed main() { IOS int t; t=1; cin>>t; while(t--){ solution(); } return 0; }

F - Coprime Present :: Panasonic Programming Contest (AtCoder Beginner Contest 195)

The main observations of this problem is if two number is not coprime, there largest prime factors of their greatest common divisors will not exceed 72.

So, we just check all possible pairs of numbers from a to b. hence there will be atmost 20 primes, upto 72. we can apply a bitmask dp approach to check which numbers will benefits us most.

Problem link

Official Editorial

Code:

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/// Bismillahir Rahmanir Rahim #include<bits/stdc++.h> using namespace std; #define ll long long #define IOS ios::sync_with_stdio(false); ///solution void solution(){ ll primes[20]= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29,31, 37, 41, 43, 47, 53, 59, 61, 67, 71}; vector<ll>dp((1<<20)); ll a,b; cin>>a>>b; dp[0]=1; for(ll i=a;i<=b;i++){ int bit=0; for(int j=0;j<20;j++){ if(i%primes[j]==0) bit|=(1<<j); } for(int j=0;j<(1<<20);j++){ if((j&bit)==0)dp[j|bit] += dp[j]; } } ll ans=0; for(int i=0;i<(1<<20);i++){ ans+=dp[i]; } cout << ans << endl; } signed main() { IOS int t; t=1; //cin>>t; while(t--){ solution(); } return 0; }

Solving a Problem With Bitmask DP Codeforces 550B

 Problem Link

In this problem the constraints is very low it can only be 15, that's why it gives a hints that it could be solve with bitmask.

Bitmask is a masking technique, by which we can have the all possible combination in 2^n complexity rather than n! complexity. To learn more about bitmask click here 

In this problem we have to calculate number of ways to choose a sum where number of elements in the sum is greater than 1 and the sum itself is at least l and at most r. And difference between maximum and minimum element which are contributed to the sum is at least x.

To do that we will check all possible subset of the set and check the requirements satisfies or not.

Code:

/// Bismillahir Rahmanir Rahim
/* Mohammad Morsalin
   Dept of ICE, NSTU
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define highest(x) numeric_limits<x>::max()
#define IOS ios::sync_with_stdio(false);
///solution
void solution(){
    ll n,l,r,x;
    cin>>n>>l>>r>>x;
    ll arr[n];
    for(int i=0;i<n;i++)cin>>arr[i];
    ll ans=0;
    for(int i=0;i<=pow(2,n);i++){
        int cnt=0;
        ll sum=0,mini=highest(ll),maxx=0;
        for(int j=0;j<n;j++){
            if(i&(1<<j)){
                cnt++;
                sum+=arr[j];
                mini=min(mini,arr[j]);
                maxx=max(maxx,arr[j]);
            }
        }
        if(sum>=l && sum<=r && cnt>=2 && maxx-mini>=x){
            ans++;
        }
    }
    cout << ans << endl;
}
signed main()
{
    IOS
    int t;
    t=1;
    //cin>>t;
    while(t--){
        solution();
    }
    return 0;
}
///Alhamdulillah