207. Course Schedule

https://leetcode.com/problems/course-schedule/description/

DFS, detect cycle. For each node, we have 3 states during DFS: 1) un-visited, 2) visiting, 3) visited.

class Solution {
public:
    bool canFinish(int n, vector<pair<int, int>>& p) {
        vector<int> visited(n, -1);     // un-visited
        vector<set<int>> v(n, set<int>());
        for (int i = 0; i < p.size(); i++) {
            v[p[i].second].insert(p[i].first);
        }
        for (int i = 0; i < n; i++)
            if (visited[i] == -1 && dfs(v, visited, i))
                return false;
        return true;
    }
    bool dfs(vector<set<int>>& v, vector<int>& visited, int n) {
        visited[n] = 1;     // visiting
        for (auto nb : v[n]) {
            if (visited[nb] == 1)
                return true;
            else if (visited[nb] == 2)
                continue;
            else {
                if (dfs(v, visited, nb))
                    return true;
            }
        }
        visited[n] = 2;     // visited
        return false;
    }
};

Topological Sort is another way.