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.
   看到这道题的想法就是画图，然后求环，有环的话就返回false。
   至于具体怎么做…想了想结果是不会做…
   答案如下，还有有点懵，之后再好好想一想！

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> matrix(numCourses);
        vector<int> in(numCourses, 0);
        int n = prerequisites.size();
        for (int i=0; i<n; i++)
            matrix[prerequisites[i].second].push_back(prerequisites[i].first);
        for (int i=0; i<numCourses; i++) {
            for (auto it=matrix[i].begin(); it!=matrix[i].end(); it++) {
                in[*it]++;
            }
        }
        for (int i=0; i<numCourses; i++) {
            int j;
            for (j=0; j<numCourses && in[j]!=0; j++);
            if (j == numCourses)
                return false;
            in[j] = -1;
            for (auto it=matrix[j].begin(); it!=matrix[j].end(); it++)
                in[*it]--;
        }
        return true;
    }
};