207. Course Schedule

最新推荐文章于 2021-12-10 13:25:51 发布

linglingconglin

最新推荐文章于 2021-12-10 13:25:51 发布

阅读量125

点赞数

CC 4.0 BY-SA版权

分类专栏：算法

本文链接：https://blog.youkuaiyun.com/u013260370/article/details/81976354

算法专栏收录该内容

55 篇文章

订阅专栏

本文介绍了一种算法，用于判断在存在先修课程约束的情况下是否能够完成所有课程。通过构建拓扑图并使用入度和出度的概念，文章提供了一个有效的方法来解决此问题。

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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.
You may assume that there are no duplicate edges in the input prerequisites.

方法1：建立拓扑图，用一个vector和一个map，分别记录每个节点的入度，出度和所指向的节点

循环n次，每次查找入度为0的点，并将该点到其他点的边都去除

判断为false条件，在某一次循环过程中找不到入度为0的点

判断为true条件，走完n次循环(表明所有的节点都能简化为入度为0)

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> indegree(numCourses,0);
        map<int,vector<int>> outdegree;
        for(auto val:prerequisites){
            indegree[val.first]++;
            outdegree[val.second].push_back(val.first);
        }
        for(int i=0,j;i<numCourses;i++){
            for(j=0;j<numCourses;j++){
                if(indegree[j]==0) break;
            }
            if(j==numCourses) return false;
            indegree[j]--;
            for(auto a:outdegree[j]) indegree[a]--;
        }
        return true;
    }
};