leetcode 207. Course Schedule 图，检查有无环，拓扑序列

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本文探讨了如何通过拓扑排序判断一组课程是否存在合理的完成顺序。在一个有向图中，节点代表课程，边表示先修关系。若图中不存在环，则所有课程均可完成；反之，则存在无法解决的课程依赖，导致部分课程无法完成。

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?

Input: 2, [[1,0]] Output: true Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.

Input: 2, [[1,0],[0,1]] Output: false Explanation: 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<vector<int>>& prerequisites) {
        int nSize = prerequisites.size();
        if(nSize == 0){
            return true;
        }
        
        int nTotal = 0;
        
        vector<int> degree(numCourses, 0);
        vector<vector<int>> G(numCourses);
        queue<int> q;
        
        // init the Graph
        for(int i = 0; i < prerequisites.size(); i++){
            degree[prerequisites[i][0]]++;
            G[prerequisites[i][1]].push_back(prerequisites[i][0]);
        }
        
        for(int i = 0; i < numCourses; i++){
            if(degree[i] == 0){
                q.push(i);      // 将第i门编号放入队列中
            }
        }
        
        while(!q.empty()){
            int t = q.front();
            q.pop();
            int edgesize = G[t].size();
            for(int i = 0; i < edgesize; i++){
                degree[G[t][i]]--;
                if(degree[G[t][i]] == 0){
                    q.push(G[t][i]);
                }
            }
            nTotal++;
        }
        
        if(nTotal != numCourses){
            return false;
        }
        return true;
    }
};