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  • 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:
    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;
        }
    };

     

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  • 原文地址:https://www.cnblogs.com/wxquare/p/6105456.html
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