Leetcode 210 Course Schedule II 拓扑排序

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is 

[0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is 

[0,1,2,3]

.
Another correct ordering is

[0,2,1,3]

.

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how
a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

课程计划1的升级版，记录修课程的顺序，与1相比，只需要在把点拿出来的时候顺便用一个容器记录一下就可以了。
关于拓扑排序，请看http://blog.csdn.net/accepthjp/article/details/62896837

class Solution {
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<int> temp(numCourses, 0);
vector<vector<int>> mp(numCourses, temp);
vector<int> in(numCourses, 0);
for(int i = 0; i < prerequisites.size(); i++)
{
mp[prerequisites[i].second][prerequisites[i].first] = 1;
in[prerequisites[i].first]++;
}
int cnt = 0;
vector<int> vis(numCourses, 0);
vector<int> res, fail;
while(++cnt <= numCourses)
{
int index = -1;
for(int i = 0; i < numCourses; i++)
if(in[i] == 0 && !vis[i])
{
vis[i] = 1;
index = i;
res.push_back(index);
break;
}
if(index == -1) return fail;
for(int i = 0; i < numCourses; i++) if(mp[index][i])in[i]--;
}
return res;
}
};