leedcode: Course Schedule

最新推荐文章于 2017-09-23 20:16:30 发布

原创最新推荐文章于 2017-09-23 20:16:30 发布 · 372 阅读

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本文探讨了如何使用深度优先搜索（DFS）算法解决课程依赖问题，通过构建有向图来判断是否能完成所有课程的学习。

There are a total of n courses you have to take, labeled from 0 ton - 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.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more abouthow a graph is represented.

这道题主要是判断有向图是否有环，用的是DFS的思想这个题目用DFS比较合适，因为DFS一条龙深入下去，如果碰到之前访问过的，就是有环

class Solution {
public:

    bool dfsVisit(vector<vector<int>*>& auxArray, int i, vector<int>& visit) {
        
        visit[i] = 1;
        if (auxArray[i] != NULL) {
            for (int j=0; j<auxArray[i]->size(); j++) {
                if (visit[auxArray[i]->at(j)] == 1) {
                    return false;
                }
                if (dfsVisit(auxArray, auxArray[i]->at(j), visit) == false) {
                    return false;
                }
            }
        }
        visit[i] = 2;
        return true;
    }
    

    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        
        if ((numCourses > 0) && (prerequisites.size() == 0))
            return true;
            
        vector<vector<int>* > auxArray(numCourses, NULL);
        
        for (int i=0; i<prerequisites.size(); i++) {
            if (auxArray[prerequisites[i].first] == NULL) {
                auxArray[prerequisites[i].first] = new vector<int>;
            }
            
            auxArray[prerequisites[i].first]->push_back(prerequisites[i].second);
        }
        
        
        vector<int> visit(numCourses, 0);
        bool ret = true;
        
        for (int i=0; i<auxArray.size(); i++) {
            if (dfsVisit(auxArray, i, visit) == false) {
                ret = false;
                break;
            }
        }
        
        for (int i=0; i<auxArray.size(); i++) {
            if (auxArray[i] != NULL)
                delete auxArray[i];
        }
        
        return ret;
    }
};