Skip to main content

Codeforces Beta Round #24 (A. Ring road)

Bismillahir Rahmanir Rahim

Thinking these case:





///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;

bool start[105],stop[105];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n,u,v,weight,sum=0,oneside=0;
    cin>>n;
    while(n--)
    {
        cin>>u>>v>>weight;
        if(stop[v] || start[u]) oneside+=weight,start[v]=stop[u]=true;
        else start[u]=stop[v]=true;
        sum+=weight;
    }
    ll otherside=sum-oneside;
    cout<<min(oneside,otherside)<<endl;
    return 0;
}
///Alhamdulillah

Problem Link: https://codeforces.com/contest/24/problem/A

Comments

Popular posts from this blog

Codeforces round 1661(B. Getting Zero)

using   second operation  the answer is maximum is  15  because  2 15 =32768 so we need to  pre-calculate all the illegible input 0 to 32768   using  second operation Then  just increment 1(using first operation)  from input check is it  minimum is not ! ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int...

Codeforces round 1676(A. Lucky?)

Just count the  first three  &  last three  number ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree   angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b)  N] + 1 = the number of digits when any number is repres...

Codeforces Round #282 (Div. 2) (B. Modular Equations)

Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include   < functional >   ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree   angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b using   namespace  std; typedef   long   long ...