LeetCode-Course Schedule-解题报告

原题链接 https://leetcode.com/problems/course-schedule/

There are a total of n courses you have to take, labeled from0 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 prerequisitepairs, 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.

如果存在环路就肯定不存在拓扑排序，开始想用并查集判断环路，然而并查集只适用于无向图。可能需要做一些修改(并没尝试过)。

其实用一次dfs就行了，尝试一次存不存在拓扑排序，判断输出节点的个数和节点总数的情况。

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<vector<int> >graph(numCourses);vector<int>out(numCourses, 0);    int num = 0;vector<bool>vis(numCourses, 0);for (auto &p : prerequisites)graph[p.second].push_back(p.first), out[p.first]++;dfs(graph, num, vis, out);return num == numCourses;}void dfs(vector<vector<int> >& g, int& num, vector<bool>& vis, vector<int>& out){if (num == g.size())return;for (int i = 0; i < g.size(); ++i){if (!vis[i] && out[i] == 0){num++;vis[i] = true;for (int j = 0; j < g[i].size(); ++j)--out[g[i][j]];dfs(g, num, vis, out);}}}};