[leetcode] 207. Course Schedule 解题报告

题目链接：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. Read more about how
a graph is represented.

click to show more hints.

Hints:

This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining
the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

思路：拓扑排序的应用．拓扑顺序就像是在一个工程中每个小项目完成有向后顺序一样，而我们的任务就是找出这个完成的先后顺利的序列．

其方法就是每次将一个入度为０的结点删除，并且将其指向的结点入度减一．

代码如下：

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph(numCourses, unordered_set<int>());//
vector<int> inDegree(numCourses, 0);//维护一个当前课程有多少先修课的数组
for(auto val:prerequisites)
{
if(graph[val.second].count(val.first) == 0)//如果当前先修课还没有被计算
{
graph[val.second].insert(val.first);//则将此课加入到先修课结点维护的集合中，代表此先修课完成之后才可以学的课
inDegree[val.first]++;//此课程的入度＋１，代表有多少先修课
}
}
int i, j;
for(i = 0; i< numCourses; i++)
{
for(j = 0; j < numCourses; j++)//寻找入度为０的结点，找到就break循环
if(inDegree[j] == 0) break;
if(j == numCourses) return false;//如果最终没有找到，说明有环
inDegree[j]--;//将这个入度为０的结点删去
for(auto val: graph[j]) //将入度为０的结点指向的结点入度都－１
inDegree[val]--;
}
return true;
}
};

参考：忘记了，sorry!