lintcode615. 课程表 拓扑排序

现在你总共有 n 门课需要选，记为 0 到 n - 1.
一些课程在修之前需要先修另外的一些课程，比如要学习课程 0 你需要先学习课程 1 ，表示为[0,1]
给定n门课以及他们的先决条件，判断是否可能完成所有课程？

样例
例1:

输入: n = 2, prerequisites = [[1,0]] 
输出: true
例2:

输入: n = 2, prerequisites = [[1,0],[0,1]] 
输出: false

一段没有想到给出的课程有10000多门而导致超时的代码

class Solution {
public:
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: true if can finish all courses or false
     */
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        // write your code here
        if(prerequisites.size()<=0) return true;
        vector<int>tmp(numCourses,0);
        queue<int> q;
        for (int i = 0; i < prerequisites.size(); i++) {
            tmp[prerequisites[i].first]++;
        }
        for (int i = 0; i < numCourses; i++) {
            if(tmp[i]==0) q.push(i);
        }
        while(!q.empty())
        {
            int num=q.front();
            q.pop();
            for (int i = 0; i < prerequisites.size(); i++) {
                if(prerequisites[i].second==num)
                {
                    tmp[prerequisites[i].first]--;
                    if(tmp[prerequisites[i].first]==0) q.push(prerequisites[i].first);
                }
            }
        }
        for (int i = 0; i < numCourses; i++) {
            if(tmp[i]!=0) return false;
        }
        return true;
    }
};

正确代码：如拓扑排序一样，先建立图的联系，然后bfs即可

class Solution {
public:
    /*
     * @param numCourses: a total of n courses
     * @param prerequisites: a list of prerequisite pairs
     * @return: true if can finish all courses or false
     */
    struct node
    {
        int indegree;
        vector<int> neighbors;
        node(int x=0):indegree(x){}
    };
    
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        // write your code here
        vector<node> graph(numCourses);
        int cnt=0;
        for (auto i : prerequisites) {
            graph[i.first].indegree++;
            graph[i.second].neighbors.push_back(i.first);
        }
        queue<node> q;
        for (int i = 0; i < numCourses; i++) {
            /* code */
            if(graph[i].indegree==0)
            {
                q.push(graph[i]);
                cnt++;
            }
        }
        while(!q.empty())
        {
            node tmp=q.front();
            q.pop();
            for (int i = 0; i < tmp.neighbors.size(); i++) {
                /* code */
                int num=tmp.neighbors[i];
                graph[num].indegree--;
                if(graph[num].indegree==0) 
                {
                    q.push(graph[num]);
                    cnt++;
                }
            }
        }
        return cnt==numCourses;
    }
};

来源：CSDN

作者：Sinb妃

链接：https://blog.csdn.net/weixin_43981315/article/details/104373326