1004. 无环图

在图论中，如果一个有向图从任意顶点出发无法经过若干条边回到该点，则这个图是一个有向无环图（Directed Acyclic Graph，DAG）. 对于一个n个节点的有向图（节点编号从0到n-1），请判断其是否为有向无环图.

图的节点数和边数均不多于100000.

请为下面的Solution类实现解决上述问题的isDAG函数，函数参数中n为图的节点数，edges是边集，edges[i]表示第i条边从edges[i].first指向edge[i].second. 如果是有向无环图返回true，否则返回false.

#include <vector>#include <iostream>#include <utility>#include <cstring>using namespace std;class Solution {public:    vector<vector<int> > V;    vector<int> visited;    bool DFS(int u) {        visited[u] = 1; // visited in this cycle        for (int i = 0; i < V[u].size(); i++) {            int v = V[u][i];            if (visited[v] == 1) {                return false;            } else if (!visited[v] && !DFS(v)){                return false;            }        }        visited[u] = 2; // have visited head vertex        return true;    }    bool isDAG(int n, vector<pair<int, int> >& edges) {        V.resize(n);        visited.resize(n);        int u, v;        for (int i = 0; i < edges.size(); i++) {            u = edges[i].first;            v = edges[i].second;            V[u].push_back(v);        }        for (int i = 0; i < n; i++)            visited[i] = 0;        for (int i = 0; i < n; i++)            if (!visited[i] && !DFS(i))                return false;        return true;    }};int main() {    pair<int, int> a[] = {{2, 1}, {0, 2}, {1,0}, {2,3}};    std::vector<pair<int, int> > edges (a, a+sizeof(a)/sizeof(a[0]));    //cout << edges[0].first << endl;    //cout << edges[1].first << endl;    Solution s;    cout << s.isDAG(5, edges);    return 0;}