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CourseSchedule.cpp
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CourseSchedule.cpp
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// Source : https://leetcode.com/problems/course-schedule/
// Author : Hao Chen
// Date : 2015-06-09
/**********************************************************************************
*
* 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.
*
* click to show more hints.
*
* Hints:
*
* - This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists,
* no topological ordering exists and therefore it will be impossible to take all courses.
*
* - Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic
* concepts of Topological Sort. (https://class.coursera.org/algo-003/lecture/52)
*
* - Topological sort could also be done via BFS. (http://en.wikipedia.org/wiki/Topological_sorting#Algorithms)
*
*
**********************************************************************************/
class Solution {
public:
bool hasCycle(int n, vector<int>& explored, vector<int>& path, map<int, vector<int>>& graph) {
for(int i=0; i<graph[n].size(); i++){
//detect the cycle
if ( path[graph[n][i]] ) return true;
//set the marker
path[graph[n][i]] = true;
if (hasCycle(graph[n][i], explored, path, graph)) {
return true;
}
//backtrace reset
path[graph[n][i]] = false;
}
//no cycle found, mark this node can finished!
explored[n] = true;
return false;
}
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
//using map to stroe the graph, it's easy to search the edge for each node
//the bool in pair means it is explored or not
map<int, vector<int>> graph;
for(int i=0; i<prerequisites.size(); i++){
graph[prerequisites[i].first].push_back( prerequisites[i].second );
}
//explored[] is used to record the node already checked!
vector<int> explored(numCourses, false);
//path[] is used to check the cycle during DFS
vector<int> path(numCourses, false);
for(int i=0; i<numCourses; i++){
if (explored[i]) continue;
if (hasCycle(i, explored, path, graph)) return false;
}
return true;
}
};