第五周拓扑排序

最新推荐文章于 2021-04-05 16:08:42 发布

aishuo_w

最新推荐文章于 2021-04-05 16:08:42 发布

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本文链接：https://blog.youkuaiyun.com/aishuo_w/article/details/66479874

本文介绍了一个典型算法问题——课程安排，通过拓扑排序判断是否能完成所有课程而不形成环。利用C++实现，构建图结构并计算每个节点的入度，最终确定所有课程是否能够顺利完成。

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算法题目： Course Schedule

算法题目描述：

There are a total of n courses you have to take, labeled from

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 abouthow a graph is represented
You may assume that there are no duplicate edges in the input prerequisites.

算法分析：

这就是典型的拓扑排序的问题了，问我们是否可以依次输出这些课并且不形成环。我们就可以用拓扑排序依次输出度为0的值，并用一个队列存入度为0的节点，然后依次出对，如果最后存在度不为0的节点，就说明存在环。应该就是这样。

算法代码（C++):

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> graph(numCourses, vector<int>(0));
        vector<int> inDegree(numCourses, 0);
        for (auto u : prerequisites) {
            graph[u[1]].push_back(u[0]);
            ++inDegree[u[0]];
        }
        queue<int> que;
        for (int i = 0; i < numCourses; ++i) {
            if (inDegree[i] == 0) que.push(i);
        }
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            for (auto v : graph[u]) {
                --inDegree[v];
                if (inDegree[v] == 0) que.push(v);
            }
        }
        for (int i = 0; i < numCourses; ++i) {
            if (inDegree[i] != 0) return false;
        }
        return true;
    }
};