Solution UVA: 161 - Traffic Lights

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Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss161 - Traffic Lights	Accepted	C++11	0.020	0.000	1830	1 mins ago

Problem:  Click 161 to read

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At the time I wrote this solution, there was almost no solution on google search, only code!

- First we need to understand the problem: it asks us to find the time at which all the lights turn green for the 2nd time (from 0 will be the time when they all turn green for the 1st time)

- To solve the problem: simply re-simulate all the lights on the time axis from 0 -> 18000 (5 hours), if there exists a moment where all the lights turn green for the 2nd time then we stop and print the result 

#include<bits/stdc++.h>

using namespace std;

int row = 0, n, n_min=1000111000;

int color[101][19000];

void f(int seconds) {

    int hours = seconds / 3600;

    int minutes = (seconds % 3600) / 60;

    seconds = seconds % 60;

    cout << setfill('0');

    cout << setw(2) << hours << ":";

    cout << setw(2) << minutes << ":";

    cout << setw(2) << seconds << "\n";

}

int main(){

    ios::sync_with_stdio(false); cin.tie(nullptr);

    while(cin >> n){

       

        if (n==0 && row==0) break;

       

        if (n==0 && row > 0){

           

            bool overtime= true;

            for(int j=n_min*2; j<=18000; j++){

                bool green= true;

               

                for(int i=1; i<=row; i++)

                if (color[i][j] != 1){

                    green= false;

                    break;  

                }

               

                if (green){

                    f(j);

                    overtime=false;

                    break;

                }

            }

            if (overtime) cout<<"Signals fail to synchronise in 5 hours\n";

            row=0;

            n_min=100011100;

            continue;

        }

       

        row++;

        n_min= min(n_min, (n==0?100011100:n));

        int red= 0;

        do{

           

            int green= red + n - 5;    

            for(int i=red; i < green; i++)

                color[row][i]= 1;

               

            int yellow = green + 5;

            for(int i=green; i<yellow; i++)

                color[row][i]= 2;

           

            red = yellow + n;

            for(int i=yellow; i<red; i++)

                color[row][i]= 3;

               

        }while(red <= 18000);

                                       

       

    }

}