leetcode：Course Schedule

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, 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.

题目解释：就是判断有向图是否有回路

思路：根据以前学数据结构上面介绍的方法就可解决，简单来说就是不断将图中入度为零的点从图中砍掉，最后图中没有点了即为非回路图，若最后还有点即为回路图。代码实现中可用一个队列保存入度为零的节点，队列为空即退出循环，并进去最后的判断。

代码：

class Solution {public:bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {int L = prerequisites.size();vector<unordered_set<int>> dppre(numCourses);for (int i = 0; i < L; i++) {dppre[prerequisites[i].first].insert(prerequisites[i].second);}queue<int> qu;for (int i = 0; i < numCourses; i++) {if (dppre[i].empty()) {qu.push(i);}}while (!qu.empty()) {int tmp = qu.front();qu.pop();for (int i = 0; i < numCourses; i++) {if (i == tmp) continue;if (dppre[i].find(tmp) != dppre[i].end()) {dppre[i].erase(tmp);if (dppre[i].empty()) {qu.push(i);}}}}for (int i = 0; i < numCourses; i++) {if (!dppre[i].empty()) return false;}return true;}};