Lecture 207, Class Schedule

力扣207,课程表

题目描述

你这个学期必须选修 numCourses 门课程,记为 0 到 numCourses - 1 .

在选修某些课程之前需要一些先修课程. 先修课程按数组 prerequisites 给出,其中 prerequisites[i] = [ai, bi] ,表示如果要学习课程 ai 则 必须 先学习课程 bi .

例如,先修课程对 [0, 1] 表示：想要学习课程 0 ,你需要先完成课程 1 .
请你判断是否可能完成所有课程的学习？如果可以,返回 true ;否则,返回 false .

输入输出样例

输入：numCourses = 2, prerequisites = [[1,0]]
输出：true
解释：总共有 2 门课程.学习课程 1 之前,你需要完成课程 0 .这是可能的.

输入：numCourses = 2, prerequisites = [[1,0],[0,1]]
输出：false
解释：总共有 2 门课程.学习课程 1 之前,你需要先完成​课程 0 ;并且学习课程 0 之前,你还应先完成课程 1 .这是不可能的.

tips

1 <= numCourses <= 105
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= ai, bi < numCourses
prerequisites[i] 中的所有课程对 互不相同

解法一,Use breadth-first search,借用队列

//Determine whether it is an infinite loop
//判断是否存在环
//Think of classes as directed edges
//Use queues to assistBFS
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
    
            //Two tables of degrees and edges are maintained
        vector<vector<int>>edges(numCourses,vector<int>());
        vector<int>indeg(numCourses);

        //Fill data into the degree table and the edge table
        for(auto info:prerequisites)
        {
    
            edges[info[1]].push_back(info[0]);
            indeg[info[0]]++;
        }

        //Use queues to help,存储入度为0的点
        queue<int>que;

        for(int i=0;i<numCourses;i++)
        {
    
            if(indeg[i]==0)
            {
    
                que.push(i);
            }
        }

        //设置标志位,Use and to check if the number of occurrences in the queue is the same as the number of courses to be taken
        int visited=0;

        while(!que.empty())
        {
    
            ++visited;
            
            int temp=que.front();
            que.pop();
            //The edge that will be associated with this dequeued vertex,入度减一
            //入度为0的,继续入栈,The loop can be circumvented here
            for(auto v:edges[temp])
            {
    
                indeg[v]--;
                if(indeg[v]==0)
                {
    
                    que.push(v);
                }
            }
        }
        return visited==numCourses;
    }

解法二,使用深度优先搜索

 //There are three states for saving,未搜索,搜索中,已完成
    bool canFinish2(int numCourses, vector<vector<int>>& prerequisites) {
    
        //initialization parameters
        vector<vector<int>>edges(numCourses,vector<int>());
        vector<int>visited(numCourses);
        bool valid=true;

        //读取数组,Import data into the edge table
        for(auto  info:prerequisites)
        {
    
            edges[info[1]].push_back(info[0]);
        }

        //Traverse each unsearched node
        for(int i=0;i<numCourses&&valid;i++)
        {
    
            //When the node is not searched,则调用dfs
            if(!visited[i])
            {
    
                dfs(i,visited,edges,valid);
            }
        }
        return valid;
    }

    void dfs(int node,vector<int>&visited,vector<vector<int>>&edges,bool &valid)
    {
    
        //Set the current node as being searched
        visited[node]=1;

        //Recursively calls the value associated with the current node
        for(int v:edges[node])
        {
    
            //0 表示存在环
            if(visited[v]==0)
            {
    
                dfs(v,visited,edges,valid);
                if(!valid)
                {
    
                    return;
                }
            }
            //1 show that there is a ring
            else if(visited[v]==1)
            {
    
                valid=false;
                return;
            }
        }
        //2 Indicates that all nodes have been completed
        visited[node]=2;
    }