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. 

解題分析：

課程的學習必須有一定的順序，也就是說學習一門課程之前可能需要先學習另外一門課程。把各個課程看做是一個有圖中的若干結點，邊的方向代表兩個課程之間的先行關係，那麼只需要判斷圖中是否存在環即可，若存在環則無法完成所有課程，否則則可以。

那麼如何判斷有向圖中是否存在環呢，常用的方法有深度優先搜索DFS，判斷是否存在反向邊，若存在則說明環。

代碼如下：

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;
}