207. Course Schedule
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- Difficulty: Medium
- Contributors: Admin
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?
解題思路：分析題意就是讓你求一個圖中是否存在迴路，最直接的辦法就是利用拓撲排序來判斷。結合在課上所講的利用dfs來進行判斷，最重要的就是一直維護每個頂點的入度序列，每次將一個入度數爲0的頂點彈出，就更新一次這個度序列
代碼如下：
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
int degree[numCourses];
vector<vector<int> >g_list;
for(int i=0;i<numCourses;i++)
{
vector<int> temp;
g_list.push_back(temp);
}
//memset(graph,0,sizeof(graph));
memset(degree,0,sizeof(numCourses));
for(int i=0;i<prerequisites.size();i++)
{
if(prerequisites[i].first==prerequisites[i].second)
return false;
else
{
degree[prerequisites[i].first]++;
g_list[prerequisites[i].second].push_back(prerequisites[i].first);
}
}
queue<int> sort;
//int first=-1;
for(int i=0;i<numCourses;i++)
{
if(degree[i]==0)
{
sort.push(i);
}
}
int count=sort.size();
//sort.push(first);
//cout<<first<<endl;
while(!sort.empty())
{
int temp=sort.front();
sort.pop();
//count++;
for(int i=0;i<g_list[temp].size();i++)
{
degree[g_list[temp][i]]--;
if(degree[g_list[temp][i]]==0)
{
count++;
sort.push(g_list[temp][i]);
}
}
}
//cout<<count<<endl;
return count==numCourses;
}
總結：拓撲排序是深度優先的很好使用，針對不同的情況可以使用圖的鄰接表或者鄰接矩陣表示，使得複雜度降低