中山大学算法课程题目详解（第四周）

问题描述：

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.

问题解决：

根据老师上课讲过的DFS算法，任意选择一个course，也就是图的任意一个点，然后进行dfs遍历，在遍历过程中如果发现有回边，那么则存在环，返回false，否则，返回true。具体的做法是，给每一个点设置一个状态，状态包括三种：未访问，正在访问，以及已访问。在dfs遍历过程中，如果发现一个正在访问的点访问另外一个访问的点，则可以判断图中存在环，则可返回false。

具体代码如下：

class Solution {
public:
Solution() {

}
vector<vector<int> > graph;
vector<int> visit;
bool dfs(int u) {
visit[u] = 1;
for (auto v : graph[u]) {
if (visit[v] == 1) {
return false;
}
if (dfs(v) == false) {
return false;
}
}
visit[u] = 2;
return true;
}
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
if (numCourses == 0 || prerequisites.empty()) {
return true;
}
graph = vector<vector<int> >(numCourses);
visit = vector<int>(numCourses, 0);
for (aut
4000
o i : prerequisites) {
graph[i.second].push_back(i.first);
}
for (int u = 0; u < numCourses; u++) {
if (visit[u] == 0 && !dfs(u)) {
return false;
}
}
return true;
}

};