算法：LeetCode207 Course Schedule

题目

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.
You may assume that there are no duplicate edges in the input prerequisites.

思路

这个题目很显然可以转换成在有向图中找环的问题，如果有环，则返回false.找环可以用dfs/bfs。这里使用dfs，在一次dfs搜索中，如果当前的搜索过程中遇到当前路径中搜索过的节点那么说明有环的存在。这里需要保存当前路径的访问状态，若没有环，则把起始节点设为之前的未访问状态，接着访问对下个节点进行dfs。

代码

class Solution {
public:
#dfs搜索，判断环
bool dfs_cycle(vector<vector<int>>& graph, int node, vector<bool>& onpath,
vector<bool>& visited)
{
if (visited[node])
return false;
onpath[node] = visited[node] = true;
for (int neigh : graph[node])
if (onpath[neigh] || dfs_cycle(graph, neigh, onpath, visited))
return true;
return onpath[node] = false;
}

bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
vector<vector<int>> graph(numCourses);
vector<bool> onpath(numCourses, false), visited(numCourses, false);
#构建图的邻接表
for (int i = 0; i < prerequisites.size(); i++)
{
graph[prerequisites[i].second].push_back(prerequisites[i].first);
}

for (int i = 0; i < numCourses; i++)
if (!visited[i] && dfs_cycle(graph, i, onpath, visited))
return false;
return true;
}

};