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.

题目非常明确的提出了先后拓扑顺序，所以采用拓扑排序来实现。在实现拓扑排序时，需要注意几点：结束的条件是没有包含入度为0的节点了；其次是出现环的时候，如(1, 0)，(0, 1)的时候该及时停止并返回false。小细节注意之后应该就很好实现了。

void GetInOutDegree(vector<pair<int, int>> &edges, vector<int> &Indegree, map<int, set<int>> &outMap,         set<int> &zeroIn)    {        unsigned len = edges.size();        pair<int, int> tmpPair;        for(unsigned i = 0; i < len; ++i)        {            tmpPair = edges[i];            if(outMap[tmpPair.first].count(tmpPair.second))                continue;            ++Indegree[tmpPair.second];            zeroIn.erase(tmpPair.second);            outMap[tmpPair.first].insert(tmpPair.second);        }    }        bool TopSort(int n, vector<int> &Indegree, map<int, set<int>> &outMap, set<int> &zeroIn,         set<int> &dealSet)    {        typedef set<int>::iterator siterator;        if(zeroIn.size() == 0)        {            if(dealSet.size() == n)  //满足拓扑排序要求                return true;            else                      //还有结点没有处理，带有环的情况                return false;        }                int tmp = *(zeroIn.begin());        zeroIn.erase(tmp);        dealSet.insert(tmp);                set<int> tmpSet = outMap[tmp];        siterator end = tmpSet.end();        for(siterator ite = tmpSet.begin(); ite != end; ++ite)        {            if(dealSet.count(*ite))  //出现环时                return false;            else            {                --Indegree[*ite];                if(Indegree[*ite] < 1)                    zeroIn.insert(*ite);            }        }                return TopSort(n, Indegree, outMap, zeroIn, dealSet);            }        bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {                int len = prerequisites.size();        if(len < 1)            return true;                if(numCourses < 2)            return true;        vector<int> Indegree(numCourses + 5, 0);        map<int, set<int>> outMap;        set<int> zeroIn;        for(int i = 0; i < numCourses; ++i)            zeroIn.insert(i);        GetInOutDegree(prerequisites, Indegree, outMap, zeroIn);        if(zeroIn.size() < 1)            return false;                 set<int> dealSet;         return TopSort(numCourses, Indegree, outMap, zeroIn, dealSet);    }