算法设计Week4 LeetCode Algorithms Problem #207 Course Schedule

本文介绍了一个算法问题,即根据课程总数及先修课程关系判断是否能完成所有课程。通过构建图结构并使用深度优先搜索(DFS)来检测环的存在,进而判断是否存在无法完成课程的情况。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目要求

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.


解题方案:
class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {

        // save the graph, make it convenient for search
        vector<unordered_set<int>> matrix(numCourses);
        for(int i = 0; i < prerequisites.size(); i++){
            matrix[prerequisites[i].second].insert(prerequisites[i].first);
        }

        vector<bool> isvisit(numCourses, false);
        vector<bool> visited(numCourses, false);

        for(int i = 0; i < numCourses; i++){
            if(!isvisit[i] && DFS(i, matrix, isvisit, visited)){
                return false;
            }
        }
        return true;
    }

    bool DFS(int course, vector<unordered_set<int>> matrix, vector<bool> isvisit, vector<bool> visited){
        isvisit[course] = true;
        visited[course] = true;
        for(auto it = matrix[course].begin(); it != matrix[course].end(); ++it){
            if(visited[*it] || DFS(*it, matrix, isvisit, visited)){
                return true;
            }
        }
        visited[course] = false;
        return false;
    }
};
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值