i want c code

first line i input N

N is the number of i need to check

second line i input $[$a$,$b$],[$b$,$c$]....$ etc $($then a$,$ b$,$ c $...$ all of them are the natural numbers.$)$

if $[0,1]$ mean if i want to check $0$ then i need to pass $1$

so for example i input line $1=4$ and line $2=[[0,1],[1,3],[2,1]]$

then

$[0,1]=$ i need to pass $1$ to pass $0$

$[1,3]=$ i need to pass $3$ to pass $1$

$[2,1]=$ i need to pass $1$ to pass $2$

so i pass $3->1->0$ or $2->$ the other$(0$ or $2)$

this example output $=$ True

but if i input line $1=2,$ line $2=[[1,0],[0,1]]$

then $[1,0]=$ i need to pass $0$ to pass $1$

$[0,1]=$ i need to pass $1$ to pass $0$

so i can't pass all of them$($because if i want to pass $1$ i need pass $0$ but i need to pass $1$ before pass $0)$

in this case output $=$ False

ofcourse $[0,0]\text{'}$s output is Flase$($because i can't pass $0)$

There are my failure code

#include

#include

#include

using namespace std;

bool canFinish$($int numCourses, vector$>$& prerequisites$)\{$

vector$>$ graph$($numCourses$)$;

vector inDegree$($numCourses$,0)$;

$//$

for $($auto& pre : prerequisites$)\{$

graph$[$pre$.$second$].$push$\_$back$($pre$.$first$)$;

inDegree$[$pre$.$first$]++$;

$\}$

queue zeroIndegree;

for $($int i $=0$; i $<$ numCourses; $++$i$)\{$

if $($inDegree$[$i$]==0)\{$

zeroIndegree.push$($i$)$;

$\}$

$\}$

int visited $=0$;

while $(!$zeroIndegree.empty$())\{$

int course $=$ zeroIndegree.front$()$;

zeroIndegree.pop$()$;

visited$++$;

for $($int next : graph$[$course$])\{$

if $(--$inDegree$[$next$]==0)\{$

zeroIndegree.push$($next$)$;

$\}$

$\}$

$\}$

$//$ True, $$ False $$

return visited $==$ numCourses;

$\}$

int main$()\{$

int N; $//$

cin $>>$ N;

vector$>$ prerequisites;

int a$,$ b;

while $($cin $>>$ a $>>$ b$)\{//$

prerequisites.push$\_$back$(\{$a$,$ b$\})$;

$\}$

cout $<<($canFinish$($N$,$ prerequisites$)?$ "True" : "False"$)<<$ endl;

return $0$;

$\}$