Course Schedule

最新推荐文章于 2019-04-15 23:20:02 发布

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本文介绍了一个算法，用于判断在存在先修课程约束的情况下，是否有可能完成所有课程。通过构建图结构并利用宽度优先搜索（BFS）策略，该算法能够有效地解决这一问题。

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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.

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        
        vector<set<int>> graph(numCourses);
        for(auto pre:prerequisites) 
            graph[pre.second].insert(pre.first);    //建立图
        
        vector<int> degrees(graph.size(),0);        //建立各个节点的入度
        
        for(auto n:graph)
        {
            for(int neigh:n)
                degrees[neigh]++;
        }
            
        for(int i=0;i<numCourses;i++)           //BFS 循环numCourses次 每次便利所有的节点，能解决就是每次都能便利到一个入度为0的节点，反之不然
        {
            int j=0;
            
            for(;j<numCourses;j++)
                if(!degrees[j]) break;
            
            if(j==numCourses) 
                return false;
            
            degrees[j]=-1;
            for(int neigh:graph[j])
                degrees[neigh]--;
     }
        return true;
        
    }
};