LeetCode：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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites. 

解题分析：

课程的学习必须有一定的顺序，也就是说学习一门课程之前可能需要先学习另外一门课程。把各个课程看做是一个有图中的若干结点，边的方向代表两个课程之间的先行关系，那么只需要判断图中是否存在环即可，若存在环则无法完成所有课程，否则则可以。

那么如何判断有向图中是否存在环呢，常用的方法有深度优先搜索DFS，判断是否存在反向边，若存在则说明环。

代码如下：

bool canFinish(int numCourses, vector<pair<int, int>>& prer) {
    vector<vector<int>> gragh(numCourses);
    vector<int> visited(numCourses, 0); // White at initialization
    for (int i = 0; i < prer.size(); i++) {
        gragh[prer[i].second].push_back(prer[i].first);
    }
    bool cycle = false;
    for (int i = 0; i < numCourses; i++) {
        if (cycle) return false;
        if (visited[i] == 0) dfs_top(i, gragh, visited, cycle);
    }
    return !cycle;
}
void dfs_top(int node, vector<vector<int>> &gragh, vector<int> &visited, bool &cycle) {
    if (visited[node] == 1) {cycle = true; return;} // cycle occurs, break the dfs chain and all return
    visited[node] = 1; //Gray, searching
    for (int i = 0; i < gragh[node].size(); i++) {
        dfs_top(gragh[node][i], gragh, visited, cycle);
        if (cycle) return; // do some pruning here
    }
    visited[node] = 2;
}