207. Course Schedule (易理解拓撲排序解法)

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.

 

題的本意就是拓撲排序，個人方法比較笨，也容易理解，就是按照拓撲排序的方法解題，464ms。

步驟：

　　（1）初始化一個大小爲numCourses的數組grap；

　　（2）將圖中個節點的入度保存到數組中，將數組中第一個入度爲0的節點找出，並刪除與它相關的邊，並將該節點在數組中的值設爲-1，其餘節點賦值爲0；

　　（3）重複（2）numCourses次，若圖中邊爲0，這返回true，不然返回false;

　　

 1 bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
 2 {
 3     int grap[numCourses]={0};
 4     for (int j=numCourses-1;j>=0;--j)
 5     {
 6         int del=-1;
 7         for (int i=0;i<prerequisites.size();i++)
 8             grap[prerequisites[i].first]++;
 9         for (int k=0;k<numCourses;k++)
10         {
11             if (grap[k]==0)
12             {
13                 del=k;
14                 break;
15             }
16         }
17         if(del==-1)
18             break;
19         vector<pair<int, int>>::iterator its;
20         for (its=prerequisites.begin();its!=prerequisites.end();)
21         {
22             if (its->second==del)
23                 its=prerequisites.erase(its);
24             else
25                 ++its;
26         }
27         grap[del]=-1;
28         for (int p=0;p<numCourses;p++)
29         {
30             if (grap[p]!=-1)
31                 grap[p]=0;
32         }
33     }
34     if(prerequisites.size()>0)
35         return false;
36     return true;
37 }

 改進：（340ms）

　　維護一個數組，數組中存放圖中各頂點入度，若頂點入度爲0，將數組中全部以該頂點爲入度的頂點入度減一。循環numCourses次，如循環中出現數組中全部頂點度都不爲0，return false.

 

 1 bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
 2 {
 3     int grap[numCourses]={0};
 4         
 5     for (int i=0;i<prerequisites.size();i++)
 6         grap[prerequisites[i].first]++;
 7     for (int j=numCourses-1;j>=0;--j)
 8     {
 9         int del=-1;
10         for (int k=0;k<numCourses;k++)
11         {
12             if (grap[k]==0)
13             {
14                 del=k;
15                 grap[k]=-1;
16                 break;
17             }
18         }
19         if(del==-1)
20             return false;
21         for (int i=0;i<prerequisites.size();i++)
22         {
23             if (prerequisites[i].second==del)
24                 grap[prerequisites[i].first]--;
25         }
26     }
27     return true;
28 }