LeetCode207. Course Schedule

题目：

https://leetcode.com/problems/course-schedule/

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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices

思路：

题意是给你n门要上的课程，编号0~n-1，再给你一个数组，存放一些课程信息，比如[0,1]表示你上课程1之前必须上过课程0。问最后你是否能完成所有的课程。
嗯，典型的拓扑排序。判断图中是否有环。
建立邻接表。
用一个队列维护入度为0的节点；
每弹出一个入度为0的节点，与之相连的节点入度减一，如果节点入度为0，加入队列中；
当队列为空后，检查所有节点的入度是否为0，若全部为0，则有向图不存在环。

代码：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> graph(numCourses,vector<int>());
        vector<int> inDegree(numCourses,0);
        int len=prerequisites.size();
        for(int i=0;i<len;i++){
            graph[prerequisites[i].first].push_back(prerequisites[i].second);
            inDegree[prerequisites[i].second]++;
        }
        queue<int> q;
        for(int i=0;i<numCourses;i++){
            if(inDegree[i]==0){
                q.push(i);
            }
        }
        int cur;
        while(!q.empty()){
            cur=q.front();
            q.pop();
            for(int i=0;i<graph[cur].size();i++){
                inDegree[graph[cur][i]]--;
                if(inDegree[graph[cur][i]]==0){
                    q.push(graph[cur][i]);
                }
            }
        }
        for(int i=0;i<numCourses;i++){
            if(inDegree[i]!=0){
                return false;
            }
        }
        return true;
    }
};