PAT暂时没有题目练习拓扑排序，以下是Leetcode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?
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.
思路1:队列方法
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
int m=prerequisites.size(),i,x=0;
vector<vector<int>>g(numCourses);
vector<int>in(numCourses,0);
for(i=0;i<m;++i){
int c1=prerequisites[i][0],c2=prerequisites[i][1];
if(c1>=0&&c2>=0&&c1<numCourses&&c2<numCourses){
g[c2].push_back(c1);
in[c1]++;
}
}
queue<int>q;
for(i=0;i<numCourses;++i)if(in[i]==0)q.push(i);
while(!q.empty()){
int u=q.front();
q.pop();x++;
for(i=0;i<g[u].size();++i){
in[g[u][i]]--;
if(in[g[u][i]]==0)q.push(g[u][i]);
}
}
if(x==numCourses)return true;
else return false;
思路2:dfs判断DAG(Directed Acyclic Gragh)
vector<vector<int>>g;vector<int>visit;
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
int m=prerequisites.size(),i,x=0;
g.resize(numCourses);
visit.resize(numCourses,0);
for(i=0;i<m;++i){
int c1=prerequisites[i][0],c2=prerequisites[i][1];
if(c1>=0&&c2>=0&&c1<numCourses&&c2<numCourses)g[c2].push_back(c1);
}
for(i=0;i<numCourses;++i)
if(!visit[i]&&!dfs(i,x))return false;
if(x==numCourses)return true;
else return false;
}
bool dfs(int& u,int& dp){
visit[u]=-1;
for(int i=0;i<g[u].size();++i){
if(visit[g[u][i]]==-1)return 0;
else if(!visit[g[u][i]]&&!dfs(g[u][i],dp))return 0;
}
visit[u]=1;dp++;return 1;
}