leetcode 207. Course Schedule

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

图论问题，有向图判断是否有环

class Solution {//判断有向图是否有环  bool is_cycle(vector<vector<int>>&adjlist, vector<int>&cnt){vector<int>visited(adjlist.size());for (int i = 0; i < adjlist.size(); i++){if (!adjlist[i].empty() && visited[i]==0){vector<vector<int>>que;vector<int>aa; aa.push_back(i); visited[i] = 1;que.push_back(aa);while (!que.empty()){vector<vector<int>>newque;for (int j = 0; j < que.size(); j++){if (!adjlist[que[j].back()].empty()){for (int k = 0; k < adjlist[que[j].back()].size(); k++){if (find(que[j].begin(), que[j].end(), adjlist[que[j].back()][k]) != que[j].end())return true;vector<int>bb = que[j];bb.push_back(adjlist[que[j].back()][k]);visited[adjlist[que[j].back()][k]] = 1;newque.push_back(bb);}}}que = newque;}}}return false;}public:bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {if (prerequisites.size() < 2)return true;vector<vector<int>>adjlist(numCourses);vector<int>cnt(numCourses);for (int i = 0; i < prerequisites.size(); i++){adjlist[prerequisites[i].second].push_back(prerequisites[i].first);cnt[prerequisites[i].first]++;}vector<int>aa;if (find(cnt.begin(), cnt.end(), 0) == cnt.end())return false;return !is_cycle(adjlist,cnt);}};

深度优先会超时，用广度优先

accepted