207. Course Schedule 图的dfs算法

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:

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

how a graph is represented.

click to show more hints.

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分析；

问题的实质是有向图判环，若存在曾经被访问过的node，则存在环。每个节点有三个状态0（白色，未访问），1（灰色，正在访问），2（黑色，访问结束）。对于无向图还需多一条当前曾被访问过的节点不是当前节点的父亲的判断。

整个算法需要先构建邻接矩阵。所以比较快。需要注意的是状态向量需要实时更新，因此在dfs中是被引用的存在。

代码：

class Solution {

public:

    void dfs(int t,vector<int>& visit,vector<vector<int>>& graph,bool& index){

        if(visit[t]==1) {index=true;return;}

         visit[t]=1;

        for(int i=0;i<graph[t].size();++i)

        {

                dfs(graph[t][i],visit,graph,index);

                if(index)

                return ;

        }

          visit[t]=2;

    }

    bool canFinish(int numCourses, vector<pair<int, int>>& pre) {

        vector<vector<int>> graph(numCourses);

        for(int i=0;i<pre.size();++i)

        graph[pre[i].first].push_back(pre[i].second);

        vector<int> visit(numCourses,0);

        bool index=false;

        for(int i=0;i<numCourses;++i)

        {

            if(index) return false;

            if(visit[i]==0) dfs(i,visit,graph,index);

        }

        return true;

    }

};