题目

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.

题目大意

你需要去学习一组课程，这些课程的标号为0-n-1。一些课程有先修课程，例如为了选修课程0，你需要去选修课程1，上述的例子可以表示为一个pair[0,1]。现在给你一个课程数和一组先修课程的列表，请你判断下你有没有可能完成所有的课程。例如2,[[1,0]]。总共就2门课给你选，你要完成课程1就得先完成课程0。所以这组课程你是可以完成的。
再如2, [[1,0],[0,1]]，总共有2门课程可以选择，为了完成课程1你得先完成课程0，并且为了完成课程0你得完成课程1，这显然是矛盾的，所以你是不可能完成的

题目分析

这题很显然，也就是大多数数据结构的书上在讲拓扑排序时引用的例子，所以很自然的联想到使用拓扑排序,代码如下:

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        /*init the graph*/
        for(int i=0;i<numCourses;i++){
            m_graph.emplace_back(vector<int>());
            visited.emplace_back(vector<bool>());
            indegree.emplace_back(0);
            for(int j=0;j<numCourses;j++){
                m_graph[i].emplace_back(0);
                visited[i].emplace_back(false);
            }
        }
        for(int i=0;i<prerequisites.size();i++){
            m_graph[prerequisites[i].second][prerequisites[i].first]=1;
            indegree[prerequisites[i].first]++;
        }
        return topSort();
    }
    bool topSort(){
        int count=0;
        stack<int> m_stack;
        for(int i=0;i<indegree.size();i++)
            if(!indegree[i]){
                m_stack.push(i);
                count++;
            }
        while(!m_stack.empty()){
            int index=m_stack.top();
            m_stack.pop();
            for(int i=0;i<m_graph.size();i++){
                if(m_graph[index][i]){
                    indegree[i]--;
                    if(indegree[i]==0)
                        m_stack.push(i),count++;
                }
            }
        }
        return count==m_graph.size();
    }
private:
    vector<vector<int>> m_graph;
    vector<vector<bool>> visited;
    vector<int> indegree;
};