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


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;
}