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.

   这个题的关键在于 就是图的思想，有没有环的存在
   方法是使用拓扑排序思想，找入度为0的点，然后删除这个点以及出边，重复这个过程，如果找不到入度为0得点就说明有环！
   
   使用indegree数组来存入度。

class Solution {public:    int findpoint(vector<int> &indegree, int num)//找入度为0的点    {        int flag = 0;        for (int i = 0; i < num; i++)        {            if(indegree[i] == 0)            {                  indegree[i]--;                return i;            }        }        return -1;    }        bool canFinish(int num, vector<pair<int, int>>& prer)     {        //形成邻接矩阵        vector<int> indegree(num, 0);        vector<vector<int>> p(num, indegree);               for (auto it : prer)        {            p[it.first][it.second] = 1;            indegree[it.second]++;        }               //就是看一个有向图是否形成环，return 0。        int taken = 0;        while (taken != num)        {            int k = findpoint(indegree, num);            if(k == -1)    //没有入度为0的点                return false;            for (int i = 0; i < num; i++)                    {                  if( p[k][i] == 1)                    indegree[i]--;            }            taken++;            }        return true;       }};