There are a total of n courses you have to take, labeled from 0 to n - 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 about how a graph is represented.
Analyse: Using BFS.
1 class Solution { 2 public: 3 bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { 4 if(prerequisites.empty()) return true; 5 6 int n = prerequisites.size(); 7 vector<vector<bool> > graph(numCourses, vector<bool>(numCourses, false)); 8 vector<int> indegree(numCourses, 0); 9 for(int i = 0; i < n; i++){ 10 if(graph[prerequisites[i].second][prerequisites[i].first] == false){//avoid duplicate pairs 11 graph[prerequisites[i].second][prerequisites[i].first] = true; 12 indegree[prerequisites[i].first]++; 13 } 14 } 15 stack<int> stk; 16 int count = 0; 17 for(int i = 0; i < numCourses; i++){ 18 if(indegree[i] == 0) 19 stk.push(i); 20 } 21 while(!stk.empty()){ 22 int source = stk.top(); 23 stk.pop(); 24 count++; 25 for(int i = 0; i < numCourses; i++){ 26 if(graph[source][i]){ 27 indegree[i]--; 28 if(indegree[i] == 0) 29 stk.push(i); 30 } 31 } 32 } 33 return count == numCourses; 34 } 35 //1. construct a graph and compute each node's indegree 36 //2. push all nodes with 0 indegree to a stack 37 //3. pop out the first node, decrease its neighbour's degree by 1 38 // push all nodes with 0 indegree to a stack and continue 39 };