Leetcode207. Course Schedule

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;
}