HackerEarth One and only flow problem solution

In this HackerEarth One and only flow problem solution You are given a network of directed graph of N vertices and M edges, where every edges' capacity is 1. An ordered pair (u,v),u!=v is good, if the maxflow is exactly 1 upon setting u as the source vertex and v as the sink vertex of the flow network. Find out if all of the ordered pairs (u,v),u!=v are good or not.

HackerEarth One and only flow problem solution.

#include  <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 5;
int n, m, tot[maxn];
bool vis[maxn], onstk[maxn], ans;
vector<int> g[maxn];

void dfs(int u) {
    vis[u] = onstk[u] = 1;
    for(int v : g[u]) {
        if(vis[v]) {
            if(!onstk[v]) {
                ans = 0;
            } else {
                tot[v]--;
                tot[u]++;
            }
        } else {
            dfs(v);
            tot[u] += tot[v];
        }
    }
    if(u != 1 and tot[u] != 1) {
        ans = 0;
    }
    onstk[u] = 0;
}

int main(int argc, char const *argv[])
{
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++)
            g[i].clear();
        for(int i = 0; i < m; i++) {
            int u, v;
            scanf("%d %d", &u, &v);
            if(u != v) {
                g[u].emplace_back(v);
            }
        }
        fill(vis, vis + n + 1, 0);
        fill(tot, tot + n + 1, 0);
        ans = 1;
        dfs(1);
        int cnt = 0;
        for(int i = 1; i <= n; i++)
            cnt += vis[i];
        if(cnt != n) {
            ans = 0;
        }
        if(ans) {
            puts("Yes");
        } else {
            puts("No");
        }
    }
    return 0;
}