使用DFS进行拓扑排序，如果可以完成的话，输出路径

进行深度优先搜索的时候，如果当前访问的点是已经被访问过的节点的话，说明出现了逆边。因此有环，无法完成拓扑排序。

如果可以完成拓扑排序，检测点序列就是一个合法的拓扑排序!

#include <iostream>#include <vector>using namespace std;class Solution {    vector<vector<int> > graph;    vector<int> visited;    vector<int> path;    void dispGraph()    {        for (int i = 0; i < graph.size(); ++i) {            cout << i << ": ";            for (int j = 0; j < graph[i].size(); ++j) {                cout << graph[i][j] << " ";            }            cout << endl;        }    }    bool dfs(int s)    {        visited[s] = 1;        for (int i = 0; i < graph[s].size(); ++i) {            int nextv = graph[s][i];            if (visited[nextv] == 0) {                if (dfs(nextv) == false)                    return false;            }            else if (visited[nextv] == 1)                return false;        }        visited[s] = 2;        path.push_back(s); // 将检测点加入到path中        return true;    }public:    bool canFinish(int numCourses, vector<pair<int, int> > &prerequisites)    {        graph = vector<vector<int> >(numCourses, vector<int>());        for (vector<pair<int, int> >::iterator it = prerequisites.begin(); it != prerequisites.end(); ++it) {            graph[it->first].push_back(it->second);        }//        dispGraph();        visited = vector<int>(numCourses, 0);        for (int i = 0; i < numCourses; ++i) {            if (visited[i] == 0 && dfs(i) == false) {                return false;            }        }        cout << "path is : ";        for (int i = 0; i < path.size(); ++i)            cout << path[i] << " ";        cout << endl;        return true;    }};int main(){    int numCourse, e;    cin >> numCourse >> e;    vector<pair<int, int> > prerequisites;    for (int i = 0; i < e; ++i) {        int next, pre;        cin >> next >> pre;        prerequisites.push_back(make_pair(next, pre));    }    cout << Solution().canFinish(numCourse, prerequisites) << endl;    return 0;}/*5 60 22 33 40 11 21 3output: path is: 4 3 2 1 015 70 22 33 40 11 21 34 3output: 0*/