leetcode Course Schedule II

题目连接

https://leetcode.com/problems/course-schedule-ii/  

Course Schedule II

Description

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$[0,2,1,3].$

拓扑排序，输出结果
Ps:我喜欢用链式前向星建图。

class Solution {
public:
	vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
		ans.clear();
		int m = prerequisites.size();
		inq = new int[numCourses + 10];
		memset(inq, 0, sizeof(int)* (numCourses + 10));
		head = new int[numCourses + 10];
		memset(head, -1, sizeof(int)* (numCourses + 10));
		G = new edge[m + 10];
		for (int i = 0; i < m; i++) {
			int u = prerequisites[i].first, v = prerequisites[i].second;
			inq[u]++;
			add_edge(v, u);
		}
		queue<int> q;
		for (int i = 0; i < numCourses; i++) { if (!inq[i])  q.push(i); }
		while (!q.empty()) {
			int u = q.front(); q.pop();
			ans.push_back(u);
			for (int i = head[u]; ~i; i = G[i].next) {
				if (--inq[G[i].to] == 0) q.push(G[i].to);
			}
		}
		delete[]G; delete[]inq, delete[]head;
		return ans.size() == numCourses ? ans : vector<int>();
	}
private:
	vector<int> ans;
	int tot, *inq, *head;
	struct edge { int to, next; }*G;
	inline void add_edge(int u, int v) {
		G[tot].to = v, G[tot].next = head[u], head[u] = tot++;
	}
};

转载于:https://www.cnblogs.com/GadyPu/p/5020731.html