leetcode 207 Course Schedule

题目描述
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.You may assume that there are no duplicate edges in the input prerequisites.

click to show more hints.
Hints:

This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.Topological sort could also be done via BFS

题意就是看看这些课程是否能找到一个顺序可以上完全部课程。其实就是一个有向图的是否存在环路的问题。可以用拓扑排序的方法，如果能找到一个完成一个拓扑排序，则有解。另外可以用一个改进的dfs算法，每个节点有三个状态，未访问，正在访问，已经访问过。对每个节点进行dfs。以下是代码：

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        if (numCourses == 0 || prerequisites.empty())            return true;        graph = vector<vector<int> >(numCourses);        vis = vector<int>(numCourses, 0); // not visit        for (auto i : prerequisites) {            graph[i.second].push_back(i.first);        }        for (int u = 0; u < numCourses; ++u) {            if (0 == vis[u] && !dfs(u))                return false;        }        return true;    }private:    vector<vector<int> > graph;    vector<int> vis;    bool dfs(int u) {        vis[u] = 1; // visiting        for (auto v : graph[u]) {            if (vis[v] == 1)                return false;            if (dfs(v) == false)                return false;        }        vis[u] = 2; // visited        return true;    };};