一、问题描述
There are a total of n courses you have to take, labeled from
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:
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
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