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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

这题类似于拓扑排序，用拓扑排序的方法可以解决

bool canFinish(int numCourses, vector<vector<int>>& prerequisites)
{
    vector<unordered_set<int>> matrix(numCourses);  //制作有向图的邻接表 
    for(int i = 0; i < prerequisites.size(); ++ i)
        matrix[prerequisites[i][1]].insert(prerequisites[i][0]); //[i][1]完成就可完成[i][0]，故用[i][1]指向[i][0] 
    
    vector<int> d(numCourses, 0); // 每个顶点入度
    for(int i = 0; i < numCourses; ++ i)
        for(auto it = matrix[i].begin(); it != matrix[i].end(); ++it)  //*it -> i 
            ++d[*it];
    
    for(int j = 0, i; j < numCourses; ++ j)
    {
        for(i = 0; i < numCourses && d[i] != 0; ++ i); //找入度为0的顶点 
        
        if(i == numCourses) //没有入度为0的顶点，即不能完成全部课程 
            return false;
        
        d[i] = -1; //去掉该顶点 
        for(auto it = matrix[i].begin(); it != matrix[i].end(); ++ it)  //被该顶点指向的点入度全减一 
            -- d[*it];
    }
    
    return true;
}