Course Schedule (graph)

最新推荐文章于 2024-09-11 04:40:53 发布

Ulricalin

最新推荐文章于 2024-09-11 04:40:53 发布

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分类专栏： graph algorithm-design 文章标签： graph

本文链接：https://blog.youkuaiyun.com/Ulricalin/article/details/78088545

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本文介绍了一种基于课程及其先修关系的课程调度算法。通过构建有向图并利用拓扑排序来判断是否能完成所有课程，解决课程安排问题。具体方法包括记录每个节点的入度、寻找入度为0的节点并更新相邻节点的入度。

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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.
You may assume that there are no duplicate edges in the input prerequisites.

FROM:
LeetCode-Algorithms-graph

Hint:
把每节课作为节点，课程之间的先导关系作为边（基础课程指向后续课程），我们可以画出一个有向图。这显然是一道判断有向图是否为有向无环图的题，只要存在拓扑序列，就是无环图。因此，这道题可以通过求解拓扑序列来获得答案。

显然我们要先找到入度为0的节点，然后把它从图中删除，同时删除相关的边，再在所得新图中继续找寻入度为0的点，重复以上过程，如果最后所有节点都被删除，删除的顺序就是一个拓扑序列，而这个图也是一个有向无环图。反之，就是存在环路，无法找到拓扑序列。

为了在删除节点时方便我们把指向后续课程的边删除，存储时我们将节点指出去的节点放在一个集合之中，每删除一个节点，就把该节点对应集合内所有节点入度-1，达到删除边的效果。

时间复杂度O(V+E)

class Solution {
public:

    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int> > graph(numCourses);
        vector<int> indegree(numCourses);//记录所有节点的入度

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

        int finishedCountNum = 0;
        queue<int> canLearnCourses;//存储入度为0的节点
        for (int i = 0; i < numCourses; i++) {
            if (!indegree[i]) {
                canLearnCourses.push(i);
            }
        }
        while(!canLearnCourses.empty()) {
            int p = canLearnCourses.front();
            canLearnCourses.pop();
            finishedCountNum++;
            for (auto j : graph[p]) {
                indegree[j]--;
                if (!indegree[j]) {
                    canLearnCourses.push(j);
                }
            }
        }
        return finishedCountNum >= numCourses;
    }
};