[leetcode] 210.Course Schedule II

题目：
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].

题意：
这道题依旧是关于拓扑排序。课程之间有依赖关系，必须完成依赖的课程，才能选该门课程。
这道题需要给出一个拓扑序列。
思路：
同201题http://blog.csdn.net/u014673347/article/details/46965275。
以上。
代码如下：

class Solution {public:    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<int> result;        if(numCourses == 0)return result;        vector<int> courses(numCourses, 0);        unordered_multimap<int,int> prereq;        for(auto p:prerequisites) {            courses[p.first]++;            prereq.insert(make_pair(p.second, p.first));        }        queue<int> que;        for(int i = 0; i < numCourses; i++) {            if(courses[i] == 0)que.push(i);        }        while(!que.empty()) {            int c = que.front();            result.push_back(c);            que.pop();            auto range = prereq.equal_range(c);            for(auto r = range.first; r != range.second; r++) {                courses[r->second]--;                if(courses[r->second] == 0)que.push(r->second);            }        }        return (result.size() == numCourses)?result:(vector<int>());    }};