LeetCode OJ 207. Course Schedule

典型的拓扑排序算法。谨以此题纪念期末机考时没AC这么简单的拓扑排序。upset……ing

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Description

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.

解题思路

拓扑排序：

1. 统计所有的定点的入度
2. 入度为0的定点入队
3. 队列定点出队时，将这个定点的所有邻接顶点入度减一，且如果入度为0时，马上入队。
4. 循环第三步，直至队列为空。

拓扑排序也可以用于判断有向图是否有环

代码

个人github代码链接

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<unordered_set<int> > adj(numCourses);        vector<int> indegree(numCourses, 0);        //存储有向图，并且计算每个点的入度        for(int i = 0; i < prerequisites.size(); i++)        {            adj[prerequisites[i].first].insert(prerequisites[i].second);            indegree[prerequisites[i].second]++;        }        //入度为0的定点入队        queue<int> q;        for(int i = 0; i < numCourses; i++){            if(indegree[i] == 0)                q.push(i);        }        int count = 0;        while(!q.empty()){            int v = q.front();            q.pop();            //输出当前入度为0的定点，并且输出计数加1            cout << v << endl;            count ++;            //所有定点v的邻接定点入度减一，并且将入度为0的定点入队            for(unordered_set<int>::iterator itr = adj[v].begin(); itr != adj[v].end(); itr++){                indegree[*itr]--;                if(indegree[*itr] == 0)                    q.push(*itr);            }        }        //输出的定点数少于全部定点数，则表示图中有回路。所以拓扑排序可以用来判断有向图有无回路        if(count < numCourses)            return false;        else            return true;    }};