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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.
题解：本质就是判断是否存在环，注意是有向图，标记数组必须标记三个变量0是未访问，1是在递归栈，2表示已从递归栈出来，所以如果访问到点标记为1说明遇到环，这里必须用邻接表表示图邻接矩阵会超时
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
const int MAXN = 1010;
vector<int> G[numCourses];
for(int i=0;i<prerequisites.size();i++){
int u = prerequisites[i].first;
int v = prerequisites[i].second;
G[u].push_back(v);
}
vector<int> color(numCourses,0);
for(int i=0;i<numCourses;i++){
if(!dfs(i,G,color)) return false;
}
return true;
}
bool dfs(int u,vector<int> G[],vector<int>& color){
if(color[u]!=0)return color[u]==2;
color[u]=1;
for(int x:G[u]){
if(color[x]==1||!dfs(x,G,color)) return false;
if(color[x]==2) continue;
}
color[u]=2;
return true;
}
};