LeetCode 207

一、问题描述

Description:

There are a total of n courses you have to take, labeled from 0 to n−1.

Some courses may have prerequisites(前提), 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.

二、解题报告

这个题目就是 “排序” 具有依赖关系的课程，看是否可以上完所有的课程，等价于所有的课程是否可以进行拓扑排序。

把每个课程看做一个顶点，课程之间的依赖关系看做一条有向边，就构成了一个有向图。而该有向图能不能拓扑排序，取决于有向图中是否存在环。这部分请看《拓扑排序的C++实现》

直接上代码：

/************************类声明************************/
class Graph
{
    int V;             // 顶点个数
    list<int> *adj;    // 邻接表
    queue<int> q;      // 维护一个入度为0的顶点的集合
    int* indegree;     // 记录每个顶点的入度
public:
    Graph(int V);                   // 构造函数
    ~Graph();                       // 析构函数
    void addEdge(int v, int w);     // 添加边
    bool topological_sort();        // 拓扑排序
};

/************************类定义************************/
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];

    indegree = new int[V];  // 入度全部初始化为0
    for(int i=0; i<V; ++i)
        indegree[i] = 0;
}

Graph::~Graph()
{
    delete [] adj;
    delete [] indegree;
}

void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); 
    ++indegree[w];
}

bool Graph::topological_sort()
{
    for(int i=0; i<V; ++i)
        if(indegree[i] == 0)
            q.push(i);         // 将所有入度为0的顶点入队

    int count = 0;             // 计数，记录当前已经输出的顶点数 
    while(!q.empty())
    {
        int v = q.front();      // 从队列中取出一个顶点
        q.pop();

        //cout << v << " ";      // 输出该顶点
        ++count;
        // 将所有v指向的顶点的入度减1，并将入度减为0的顶点入栈
        list<int>::iterator beg = adj[v].begin();
        for( ; beg!=adj[v].end(); ++beg)
            if(!(--indegree[*beg]))
                q.push(*beg);   // 若入度为0，则入栈
    }

    if(count < V)
        return false;           // 没有输出全部顶点，有向图中有回路
    else
        return true;            // 拓扑排序成功
}


class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {

        Graph g(numCourses);  // 创建图
        for(int i=0; i<prerequisites.size(); ++i)
            g.addEdge(prerequisites[i][1], prerequisites[i][0]);

        return g.topological_sort();
    }
};

LeetCode答案源代码：https://github.com/SongLee24/LeetCode