[Leetcode] 207. Course Schedule 解题报告

题目：

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.

click to show more hints.

Hints:

1. 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.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

思路：

这是一道拓扑排序的题目。我们的思路是首先根据先修课程的要求构建出来一张图，并且记录每个图的入度（indegree）。然后每次取出一个入度为0的顶点，作为当前可以选修的课程，同时更新和该课程构成先修关系的其它课程的入度值。如果我们做了n遍，每次都可以选出入度为0的顶点，则说明该图存在一个合法的拓扑排序，所以返回true。否则如果在某一次循环中，发现已经无法找到入度为0的顶点，这说明我们遇到了cycle，该图没有合法的拓扑排序，所以返回false。

我们下面实现的拓扑排序是基于BFS的。提示说用DFS也可以实现，我也记得算法课上老师讲了用DFS实现拓扑排序的思路，不过早已经还给老师了。。。有兴趣的读者可以戳提示中给出的视频链接。

代码：

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<unordered_set<int>> graph(numCourses, unordered_set<int>());        vector<int> indegree(numCourses, 0);        for(auto pre : prerequisites) {                     // initialize the graph            if(graph[pre.second].count(pre.first) == 0) {                graph[pre.second].insert(pre.first);                indegree[pre.first]++;            }        }        for(int i = 0, j = 0; i < numCourses; i++) {            for(j = 0; j < numCourses; j++) {               // find the course that can be taken now                if(indegree[j] == 0) {                    break;                }            }            if(j == numCourses) {                 return false;            }            --indegree[j];                                  // update the indegree related to j            for(auto p : graph[j]) {                --indegree[p];            }        }        return true;    }};