Course Schedule

最新推荐文章于 2022-02-16 14:03:41 发布

转载最新推荐文章于 2022-02-16 14:03:41 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/wxquare/p/6105456.html

探讨了课程安排问题，即判断是否能完成所有课程，若存在先修课程则需先完成。通过将问题转化为判断有向图是否存在环的问题，并采用拓扑排序的方法解决。介绍了如何构建图和进行拓扑排序的具体实现。

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.

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.

本题最终转化为求有向图中是否有存在环，转化为拓扑序问题

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> graph(numCourses);
        vector<int> in_degree(numCourses, 0);

        for (auto p : prerequisites) {
            graph[p.second].push_back(p.first);
            in_degree[p.first]++;
        }

        queue<int> q;
        int cnt = 0;
        for (int i = 0; i < numCourses; i++) {
            if (in_degree[i] == 0)
                q.push(i);
        }
        while (!q.empty()) {
            int cur = q.front();
            q.pop();
            for (auto it = graph[cur].begin(); it != graph[cur].end(); it++) {
                if (--in_degree[*it] == 0)
                    q.push(*it);
            }
        }

        for (int i = 0; i < numCourses; i++) {
            if (in_degree[i] != 0)
                return false;
        }
        return true;
    }
};