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?
Example 1:
Input: 2, [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]] Output: false Explanation: 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：
这道题其实就是看学习的课程之间时候存在环路，如果存在环路那么就不能完成学习。我们这里使用深度优先搜索判断时候存在环路：如果从一个点出发开始深度优先遍历，那么之后再次访问到该点，就说明有环出现了。
代码如下：
class Solution {
public:
vector<vector<int>> admat; // 存储邻接矩阵
vector<int> states; //0表示没有被访问过
void genAdmat(const int &nodeNum, const vector<pair<int, int>> &sites){
admat = vector<vector<int>>(nodeNum, vector<int>(nodeNum, 0));
for(pair<int, int> site:sites){
admat[site.first][site.second] = 1;
}
}
// 深度优先遍历
bool hasloop = false;
void DFS(int node){
states[node] = 1; //1表示是初始访问节点
int nodeNum = int(states.size());
for(int i = 0;i<nodeNum;i++){
if(admat[node][i] != 0){
if(states[i] == 1) {hasloop = true; break;} //发现已经遍历过的节点，说明有环
else if(states[i] == -1) continue;
else DFS(i);
}
}
states[node] = -1; //表示已经被访问过了
}
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
genAdmat(numCourses, prerequisites);
states = vector<int>(numCourses, 0);
for(int i = 0;i<numCourses;i++){
if(states[i] == -1) continue;
DFS(i);
if(hasloop) return false; //如果存在环则不能修完
}
return true;
}
};