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.

找是否存在环，用深搜。不难，但因为数组越界debug了好久。吸取到一个教训，为方便debug以后在可能存在越界的地方加上断言。

class Solution {public:    int *visit;    bool isCircle;    void isExistCircle(vector<vector<int> > &a,int i)    {       if(visit[i] == 1 || isCircle) return;       visit[i] = -1;       for(int j = 0;j<a[i].size();++j)       {           if(visit[a[i][j]] == -1){ isCircle = true; return; }           isExistCircle(a,a[i][j]);       }       visit[i] = 1;    }        bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {                if(numCourses == 0) return true;                vector<vector<int> > a(numCourses,vector<int>());        visit = new int[numCourses];        memset(visit,0,numCourses*sizeof(int));        isCircle = false;                for(int i=0;i<prerequisites.size();++i)        {            a[prerequisites[i].first].push_back(prerequisites[i].second);        }                for(int i=0;i<numCourses;++i) isExistCircle(a,i);                delete []visit;        return !isCircle;    }};