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leetcode/algorithms/cpp/courseSchedule/CourseSchedule.cpp at master · wcoder0/leetcode · GitHub
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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
;
}
};
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