leetcode [Graph] No.207 Course Schedule

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题目描述

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?

Example1:

Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

Example2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

解题思路

该问题中，我们可以根据课程学习的先后顺序构建一张有向图，其中，若课程a必须在修完课程b后才可以学习，那么该有向图中有一条边，由b指向a。在构建完有向图之后，该问题便变成了有向图中是否存在一个环，若存在环，便无法学习完所有课程，反之亦然。我们可以使用拓扑排序算法简单、高效的判断一个有向图中是否存在一个环，时间复杂度为 $O(n+e)$，其中 $n$表示顶点个数， $e$表示边的个数。

拓扑排序步骤

1. 遍历图中所有的边，记录下每个定点的入度。
2. 遍历所有定点的入度，将入度为0的顶点压入一个队列。
3. 当队列不为空时，抛出队列中第一个元素，否则跳到步骤5.
4. 将所有从抛出点可到达的顶点的入度减一，若某个点的入度为0，则加入队列尾部。循环辉步骤3。
5. 检查所有定点的入度，若存在一个顶点入度不为0，则该有向图存在环，否则不存在。

代码：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        //新建队列
        queue<int> q;
        //初始化有向图
	    bool map[numCourses][numCourses] = {};
	    //记录顶点入度的数组
        int in[numCourses] = {};
        //根据输入构建有向图
        for (auto iter : prerequisites) {
        	map[iter.second][iter.first] = true;
        	in[iter.first]++;
		}
		
		//入度为0的点入队
		for (int i = 0; i < numCourses; i++) {
			if (in[i] == 0) {
				q.push(i);
			}
		}
	
		//循环求解
		while(!q.empty()) {
			int course = q.front();
			q.pop();
			for (int i = 0; i < numCourses; i++) {
				if (map[course][i]) {
					if (--in[i] == 0) {
						q.push(i);
					}
				}
			}
		}
		
		//检查是否还存在入度不为0的点
		for (int i = 0; i < numCourses; i++) {
			if (in[i] != 0) return false;
		}
		return true;
    }
};

运行结果：