[Leetcode]Course Schedule

There are a total of n courses you have to take, labeled from 0 ton - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair:[0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more abouthow a graph is represented.

class Solution {public:    /*algorithm:bfs        https://en.wikipedia.org/wiki/Topological_sorting    */    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {         queue<int>S;         int L;         //count in degree for each node          vector<int>in(numCourses,0);         for(int i = 0;i < prerequisites.size();i++)in[prerequisites[i].first]++;         //store in degree 0 nodes to S         for(int i = 0;i < in.size();i++){             if(!in[i])S.push(i);         }         L = S.size();         //topological sort         while(!S.empty()){             int n = S.front();S.pop();             for(int i = 0;i < prerequisites.size();i++){                 if(prerequisites[i].second == n){                     int m = prerequisites[i].first;                     in[m]--;                     if(!in[m]){                        S.push(m);                        L++;                     }                 }             }         }         return L == numCourses;    }};

class Solution {public:    /*algorithm:dfs        1)construct adjacent list for graph        2)do dfs to detect loop    */    vector<vector<int> >buildAdjacetList(vector<pair<int, int>>& edges,int n){        vector<vector<int> >result(n,vector<int>());        //edge: v->w        for(int i = 0;i < edges.size();i++){            int v = edges[i].second,w = edges[i].first;            result[v].push_back(w);        }        return result;    }    bool dfs(vector<vector<int> >&graph,int v,vector<int>&visit){         if(visit[v] == -1)return false;         if(visit[v] == 1)return true;         visit[v] = -1;         for(int i = 0;i < graph[v].size();i++)            if(!dfs(graph,graph[v][i],visit))return false;         visit[v] = 1;         return true;    }    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {         vector<vector<int> >graph = buildAdjacetList(prerequisites,numCourses);         //0: 1  1:<>         vector<int>visit(numCourses,0);         for(int i = 0;i < numCourses;i++){            if(!dfs(graph,i,visit))return false;         }         return true;    }};