算法作業_42(2017.6.27第十九周)

207. 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.

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.

DFS的解法，需要建立有向圖，還是用二維數組來建立，我們像現在需要一個一維數組visit來記錄訪問狀態，大體思路是，先建立好有向圖，然後從第一個門課開始，找其可構成哪門課，暫時將當前課程標記爲已訪問，然後對新得到的課程調用DFS遞歸，直到出現新的課程已經訪問過了，則返回false，沒有衝突的話返回true，然後把標記爲已訪問的課程改爲未訪問。代碼如下：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector<int>> g (numCourses,vector<int>(0));
        vector<int> visit (numCourses,0);
        for(auto a :prerequisites){
            g[a.first].push_back(a.second);
        }
        for(int i = 0 ; i <numCourses ; i++){
            if(!DFS(g,visit,i)){
                return false;
            }
        }
        return true;
    }
private:
    bool DFS(vector<vector<int>> &g ,vector<int> &visit ,int i ){
        if(visit[i] == -1) return false;
        if(visit[i] == 1) return true;
        visit[i] = -1 ;
        for(auto a : g[i]){
            if(!DFS(g,visit,a)) return false;
        }
        visit[i] = 1 ;
        return true ; 
    }
};