LeetCode -- Course Schedule

题目描述：


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.


选课问题，需要选n门课，已知依赖关系为二维数组[{1,0},{0,2},{x,y}..]，表示课程0依赖课程1（要选0必须先选1），课程0依赖课程2，课程x依赖课程y。课程id的取值范围是[0,n)。


思路：
每个课程看作一个节点，依赖关系数组其实是一个图结构。对于[{0,1},{2,3}]这个依赖数组来说，其实相当于图中的0->1 , 2->3，只要不存在环，选课操作就是可以继续的。
1. 使用referencing[0...n-1]来存当前课程所依赖其他课程的数量
2. 使用标记数组selections[0...n-1]来标记课程selections[i]是否已被选
3. 每次遍历referencing数组，找出依赖数为0并且还没有被选的课，这些课程id将被选择并放入集合S；然后遍历S，在依赖关系图中找到谁依赖了S[i]，拿到这个课程id: course，然后用于更新引用数组：referencing[course]--，重复步骤1。
4. 直到referencing中没有元素0或选够了n门课为止。




实现代码：

public class Solution {    public bool CanFinish(int numCourses, int[,] prerequisites)     {        var row = prerequisites.GetLength(0);    if(row < 2){    return true;    }        var referencing = new int[numCourses];    var selections = new bool[numCourses];    for(var i = 0;i < row; i++)    {    referencing[prerequisites[i,0]] ++;    }        var c = 0;    while(c < numCourses){        // find out which courses not referencing others    var selected = new List<int>();    for(var i = 0;i < referencing.Length; i++){    if(referencing[i] == 0 && !selections[i])    {    // select this course    c++;    selections[i] = true;    selected.Add(i);    }    }        if(selected.Count == 0)    {    break;    }        // find out which ones referencing this course then update referencing counter    foreach(var s in selected){    for(var j = 0; j < row; j++){    if(prerequisites[j,1] == s){    referencing[prerequisites[j,0]] --;    }    }    }    }        return c >= numCourses;        }}