Course Schedule -- leetcode

标签：leetcode   拓扑排序   广度优先   

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.

基本思路：拓扑排序

课程之间有依赖。

1.找出那些前置条件已经具备的课程进行学习。

2.学完该课程后，更新依赖该门课程的其他课程。

3. 重复步骤1和2.

算法中，维持一个度数数组，或者称前置课程数组degree。表示要学习该课程时，尚需要学习的前置课程数。

如果其值为0，则该课程可以开始学习了。

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> degree(numCourses);
        unordered_set<int> finished_courses;
        for (auto p: prerequisites)
            ++degree[p.first];
            
        for (int i=0; i<numCourses; i++) {
            if (!degree[i])
                finished_courses.insert(i);
        }
        
        int count = 0;
        while (!finished_courses.empty()) {
            int course = *finished_courses.begin();
            finished_courses.erase(finished_courses.begin());
            for (auto p: prerequisites) {
                if (p.second == course && !--degree[p.first])
                    finished_courses.insert(p.first);
            }
            ++count;
        }
        
        return count == numCourses;
    }
};

Course Schedule -- leetcode

标签：leetcode   拓扑排序   广度优先   

原文地址：http://blog.csdn.net/elton_xiao/article/details/47207927