Isomorphic graph parallel code implementation using openmp - c++

Question

I am trying to implement a parallel code for graph isomorphism(checking if 2 graphs are isomorphic or not) using openmp. I have the sequential code below, written in C++ which works only for regular graphs.
I have taken the input from a text file (input.txt) which contains number of nodes and 2 adjacency matrices representing 2 graphs. The output.txt file indicates whether the graphs are isomorphic or NON isomorphic along with the execution time.
I have used is_regular() function to check if a graph is regular graph or not.
Can someone please help me parallelize this code using openmp.
#include <time.h>
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <vector>
#include <string>
using namespace std;
bool is_regular(vector<string> &adj) {
int nv = adj.size();
int cnt = 0;
for (int j = 0; j < nv; ++j)
if (adj[0][j] == '1') ++cnt;
for (int i = 1; i < nv; ++i) {
int sch = 0;
for (int j = 0; j < nv; ++j)
if (adj[i][j] == '1') ++sch;
if (sch != cnt) return false;
}
return true;
}
vector<vector<int> > foo1(vector<string> &adj, char R) {
int nv = adj.size();
vector<vector<int> > res;
for (int v1 = 0; v1 < nv; ++v1) {
vector<int> tmp;
vector<bool> b(nv);
vector<int> p, q;
p.push_back(v1);
b[v1] = true;
while (!p.empty()) {
q.clear();
for (size_t i = 0; i < p.size(); ++i) {
for (int j = 0; j < nv; ++j)
{
if (!b[j] && adj[p[i]][j] == R) {
b[j] = true;
q.push_back(j);
}
}
}
if (q.empty()) break;
vector<int> d(p.size());
vector<int> e(q.size());
for (size_t i = 0; i < p.size(); ++i) {
for (size_t j = 0; j < q.size(); ++j) {
if (adj[p[i]][q[j]] == R) {++d[i]; --e[j];}
}
}
sort(d.begin(), d.end());
sort(e.begin(), e.end());
copy(d.begin(), d.end(), back_inserter(tmp));
copy(e.begin(), e.end(), back_inserter(tmp));
p.swap(q);
}
res.push_back(tmp);
}
sort(res.begin(), res.end());
return res;
}
vector<vector<int> > foo2(vector<string> &adj, char R) {
int nv = adj.size();
vector<vector<int> > res;
for (int v1 = 0; v1 < nv; ++v1) {
for (int v2 = v1 + 1; v2 < nv; ++v2) {
vector<int> tmp;
vector<bool> b(nv);
vector<int> p, q;
p.push_back(v1);
p.push_back(v2);
b[v1] = true;
b[v2] = true;
while (!p.empty()) {
q.clear();
for (size_t i = 0; i < p.size(); ++i) {
for (int j = 0; j < nv; ++j) {
if (!b[j] && adj[p[i]][j] == R) {
b[j] = true;
q.push_back(j);
}
}
}
if (q.empty()) break;
vector<int> d(p.size());
vector<int> e(q.size());
for (size_t i = 0; i < p.size(); ++i) {
for (size_t j = 0; j < q.size(); ++j) {
if (adj[p[i]][q[j]] == R) {++d[i]; --e[j];}
}
}
sort(d.begin(), d.end());
sort(e.begin(), e.end());
copy(d.begin(), d.end(), back_inserter(tmp));
copy(e.begin(), e.end(), back_inserter(tmp));
p.swap(q);
}
res.push_back(tmp);
}
}
sort(res.begin(), res.end());
return res;
}
vector<vector<int> > foo3(vector<string> &adj, char R) {
int nv = adj.size();
vector<vector<int> > res;
for (int v1 = 0; v1 < nv; ++v1) {
for (int v2 = v1 + 1; v2 < nv; ++v2) {
for (int v3 = v2 + 1; v3 < nv; ++v3) {
vector<int> tmp;
vector<bool> b(nv);
vector<int> p, q;
p.push_back(v1);
p.push_back(v2);
p.push_back(v3);
b[v1] = true;
b[v2] = true;
b[v3] = true;
while (!p.empty()) {
q.clear();
for (size_t i = 0; i < p.size(); ++i) {
for (int j = 0; j < nv; ++j) {
if (!b[j] && adj[p[i]][j] == R) {
b[j] = true;
q.push_back(j);
}
}
}
if (q.empty()) break;
vector<int> d(p.size());
vector<int> e(q.size());
for (size_t i = 0; i < p.size(); ++i) {
for (size_t j = 0; j < q.size(); ++j) {
if (adj[p[i]][q[j]] == R) {++d[i]; --e[j];}
}
}
sort(d.begin(), d.end());
sort(e.begin(), e.end());
copy(d.begin(), d.end(), back_inserter(tmp));
copy(e.begin(), e.end(), back_inserter(tmp));
p.swap(q);
}
res.push_back(tmp);
}
}
}
sort(res.begin(), res.end());
return res;
}
int main() {
int nt = 1;
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int n;
int tc = 0;
while (cin >> n) {
vector<string> adj1(n);
vector<string> adj2(n);
for (int i = 0; i < n; ++i) cin >> adj1[i];
for (int i = 0; i < n; ++i) cin >> adj2[i];
clock_t start = clock();
string answer = "NON-isomorphic";
if (!is_regular(adj1) || !is_regular(adj2)) {
answer = "Non regular graph";
goto mmm;
}
if (foo1(adj1,'1') != foo1(adj2,'1') || foo1(adj1,'0') != foo1(adj2,'0')) {
answer += " * ";
goto mmm;
}
if (foo2(adj1,'1') != foo2(adj2,'1') || foo2(adj1,'0') != foo2(adj2,'0')) {
answer += " ** ";
goto mmm;
}
if (foo3(adj1,'1') != foo3(adj2,'1') || foo3(adj1,'0') != foo3(adj2,'0')) {
answer += " ***";
goto mmm;
}
answer = "isomorphic";
mmm:
++tc;
printf("%d) n = %d %s", tc, n, answer.c_str());
printf(" %f\n", (double)(clock() - start) / CLOCKS_PER_SEC);
}
return 0;
}