207. Course Schedule

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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.

题目大意

前提：有部分课程需要以另一门课程为前提，如果存在两门课互为前提，则该课程信息不可完成；如果不存在，则该课程信息可完成。
输入课程信息，判断该课程信息能否完成。

解题思路

这是一个拓扑排序问题。
遍历有向图中的节点，每次找到一个入度为0的节点，将其push进队列。然后利用BFS，取队列的头(并将其pop掉)，将该节点所指向其他节点的边都去掉（即，将其指向的节点入度减一），如果存在节点入度减一的值为零，则将其push入队列，重复这个过程，直到队列为空。
最后检查送队列中被pop掉的节点总数，如果总数等于课程数量，则说明该有向图无环，课程信息可完成；如果总数小于课程数量，则说明该有向图存在环，课程信息不可完成。

算法复杂度

O(|V+E|)

代码实现

class Solution {public:  bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {    vector<int> inDegreeMap(numCourses);    vector<pair<int,int>>::iterator iter;    // compute the in-degree of all points    for (iter = prerequisites.begin(); iter != prerequisites.end(); iter++) {      ++inDegreeMap[iter->second];    }    // push back the 0-inDegree point to the queue    queue<int> que;    for (int i = 0; i < numCourses; ++i) {      if (inDegreeMap[i] == 0) { que.push(i); }    }    // BFS    int count = que.size();    while (!que.empty()) {      int u = que.front();      que.pop();      for (iter = prerequisites.begin(); iter != prerequisites.end(); iter++) {        if (u == iter->first) {          // find the next position of u          // and determine whether its (in-degree - 1) equals 0          int uLast = iter->second;          if (--inDegreeMap[uLast] == 0) {            que.push(uLast);            ++count;          }        }      }    }    return count == numCourses;  }};