Course Schedule

原题：来自leetcode的https://leetcode.com/problems/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, 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.

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> visited(numCourses, false);//记录每个节点是否已经被访问过，访问过，且没有环，就改成2，使用过程中，将其置1
        vector<vector<int>> graph(numCourses);
        for(int i=0; i<prerequisites.size(); i++){//创建拓扑排序的说结构，这里选择动态二维数组实现的数据结构
            graph[prerequisites[i].second].push_back(prerequisites[i].first);
        }
        
        for(int i=0; i<numCourses; i++){
            if(visited[i]==0 && !judge(visited, i, graph))
                return false;
        }
        return true;
    }
    
    bool judge(vector<int> &visited, int idxCourses, vector<vector<int>> & graph){//使用dfs实现是否有环的判断
        if(visited[idxCourses]==1)//如果发现为1，表示从idxCourses出发又回到了idxCourses
            return false;
        visited[idxCourses]=1;//改变颜色为1
        for(auto it=graph[idxCourses].begin(); it!=graph[idxCourses].end(); it++){
            if(!judge(visited, *it, graph))
                return false;
        }
        visited[idxCourses]=2;//表示从当前节点除法不会差生环
        return true;
    }
};