LeetCode 207 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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

这道题其实就是检查图有没有环，检查图有没有环比较常用的方法就是使用拓扑排序，拓扑排序的思路就是不断从图中取出入度为0的点，直到不能没有入度为0的点，这样操作之后如果可以将图中所有的点全部取出，那么说明这个图是可以进行拓扑排序的，本题的具体的方法如下流程：
1. 找到入度为0的节点，入队
2. 队列是否为空
2.1 为空则跳出循环
2.2 不为空则取出一个节点，取出的记录数+1，作为当前点，遍历该当前点的所有邻接点，并将所有节点的入度均-1，表示取出了当前点，如果有节点的入度-1之后变为0，那么将其入队。
3. 取出的节点是否等于总的节点数
3.1 等于则说明没有环
3.2不等于则说明有环

具体代码如下：

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {         vector<vector<int> > relation(numCourses);         int size = prerequisites.size();         vector<int> indegree(numCourses);         int outnum = 0;         for(int i = 0; i < size; i++)         {             pair<int, int> pairtmp = prerequisites[i];             relation[pairtmp.first].push_back(pairtmp.second);             indegree[pairtmp.second]++;         }         stack<int> wait;         for(int i = 0; i < numCourses; i++)         {             if(indegree[i] == 0)                wait.push(i);         }         while(!wait.empty())         {             int cur = wait.top();             wait.pop();             outnum++;             for(int i = 0; i < relation[cur].size(); i++)             {                 int adject = relation[cur][i];                 indegree[adject]--;                 if(indegree[adject] == 0)                    wait.push(adject);             }         }         if(outnum == numCourses)            return true;        else            return false;    }};