You also need to track the maximum time taken by any single person. Coz edge case may be that a person takes more time than anybody else, and goes beyond sum(exi)+min(ei). Once you have the maximum time taken by any one person, the answer will be max(sum(exi)+min(ei), max_time)

im getting WA on most of the cases. can anyone help whats the problem i did here ?

Hi there! There is no sorting done in my C++ solution but it is not able to pass only 1 test case. This is the code.

#include <iostream>

For everyone got TLE (the time limits are very strict).

I implemented the dijkstra by priority_queue and got TLE with testcase 7 and 8. After that, I added an array visited[] to ...

not able to understand the can someone make me understand why dijkstra is use

Thanks a lot for the link!

#include<bits/stdc++.h>

using namespace std;

WA for 7 test cases. I tried it without k -= k > 0 ? 1 : 0 but it always seems that I have to decrease k by 1 to get the right answer. Please help!

Here is my solution: Solution: 28918761

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int ansf[1000005];

#include <bits/stdc++.h>
#include <list>
long long N;
std::vector <long long> graphs[100010];
bool vis[100010];

arr=list(map(int,input().split()))
n=arr[0]
graph=[[]for i in range(n+1)]
cost=[]
for i in range(1,n+1):

///just get the smallest child for every node than loop over them to get the answer 
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<int,int>

This might help: International Olympiad in Informatics: Path to Gold

This playlist is mostly based on IARCS syllabus and previous years' papers' trends.

Do we have to assume that the number of roads before and after the reform are same?

Wrong answer on test case 10.

Please, help me. Here's my code:

My logic is that the graph may be a collection of disconnected graphs. So a disconncted graph has a cycle=0 isolated cities, otherwise 1 isolated cities.

#include <bits/stdc++.h>
 
using namespace std;
long long n,m,x,y;
vector <long long> graph[1000000];
bool vis[1000000];
int main()
{
    long long cycle (long long,long long);
    cin>>n>>m;
 
    for(long long i = 0 ; i < m ; i++)
    {
        cin>>x>>y;
        graph[x].push_back(y);
        graph[y].push_back(x);
    }
 
 
    for(long long i = 1 ; i <= n ; i++ )       //This loops helps in navigating through different disconnected graphs
    {
         x=0;                                   // x is used inside cycle()
        if( !vis[ i ] ){   
            vis[1]=1;
            isolated += ( !cycle( i , -1 ) ) ; //if it is a cycle add 0, else add 1
        }
    }
 
    cout<<isolated;
 
    return 0;
}
 
long long cycle(long long cur, long long prev)
{
    long long x=0;
 
    for(unsigned long long i = 0 ; i < graph[cur].size(); i++ )
    { 
        if( !vis[ graph[ cur ][ i ] ] )
        {
            vis[ graph[ cur ][ i ] ] = 1;
            x+=cycle( graph[ cur ][ i ] , cur );
        }
        else if( prev != graph[ cur ][ i ] )
             {
                 x++;   //It has a cycle (so we will add 0 to isolated.
             }
    }
    return x;
}
 

n,m=map(int,input().split())
 
graph=[[]for i in range(n)]
for i in range(m):
  a,b=map(int,input().split())