Leetcode编程练习：课程编排

题目原文：（id=210)

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].

简单来说，就是输入一些课程的先后修的关系，要求找到一种可行的修课程的顺序，是有向无环图拓扑排序的一个典型运用。首先在输入时把边表转化为邻接矩阵，同时求出（初始化）各个顶点的入度（也就是每一门课程参与前要修完的课程种类数），如果没有入度为0的点，则肯定没有合法的解，否则将入度为0的顶点存储在一个队列中。然后将队列中顶点逐个取出，把所有当前顶点指向的顶点的入度减1，如果它们变成0则放到队列当中，直到所有元素都出队。顶点出队的顺序显然就是图的一个拓扑排序。这个算法时间复杂度为O(E)

vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        int n = numCourses;
		vector<vector<int> > adj(n);
		vector<int> indegree(n, 0);
		for (int i = 0;i < prerequisites.size();i++) {
			adj[prerequisites[i].second].push_back(prerequisites[i].first);
			indegree[prerequisites[i].first]++;
		}
		vector<int> seq;
		queue<int> open;
		for (int i = 0;i < n;i++) {
			if (indegree[i] == 0) {
				open.push(i);
				seq.push_back(i);
			}
		}
		while (open.size()) {
			int pos = open.front();
			open.pop();
			for (int i = 0;i < adj[pos].size();i++) {
				int nw = adj[pos][i];
				indegree[nw]--;
				if (indegree[nw] == 0) {
					open.push(nw);
					seq.push_back(nw);
				}
			}
		}
		if (seq.size() == n) return seq;
		else return vector<int>();
    }