leetcode-Course Schedule

leetcode-Course Schedule

題目：leetcode

Course Schedule

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.

分析：

本題目可轉化為 “判斷有向圖中是否有環”。

1、把數組prerequisites中的內容放進哈希表table，table的key是課程編號，value是一個數組，記錄了key代表的課程的約束條件（即上完value中的課程，才能上key的課程）。如果每一門課程當成平面上的一個點，那麼key和value中的每一門課程，可以分別連成一條有向路徑。

2、利用回溯法，如果沒有環，則把相應的“路徑”刪除，直到把有向圖刪光。若有環，則整個程序返回false。

class Solution {
public:
    bool canFinish(int numCourses, vector>& prerequisites) {
        if(prerequisites.size()<=1)
            return true;
        
        unordered_map> table;
        for(auto &i:prerequisites)
        {
            table[i[0]].push_back(i[1]);
        }
       vector path;
        while(!table.empty())
        {
            auto it=table.begin();
            path.push_back(it->first);
            if(HasLoop(table,it->first,path))
                return false;
            path.pop_back();
        }
        return true;
    }
    //判斷是否有環，有環的話返回真，否則返回假
    bool HasLoop( unordered_map> &table,const int &begin,vector &path)
    {
       if(table.count(begin)==0)
         return false;
       while(!table[begin].empty())
       {
           int temp=table[begin].back();
           if(find(path.begin(),path.end(),temp)!=path.end())
                return true;
            path.push_back(temp);
           if(HasLoop(table,temp,path))
                return true;
            path.pop_back();
            table[begin].pop_back();
       }
       table.erase(begin);
       return false;
    }
    
};