算法设计与应用基础： 第五周（1）

207. Course Schedule

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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?

解题思路：分析题意就是让你求一个图中是否存在回路，最直接的办法就是利用拓扑排序来判断。结合在课上所讲的利用dfs来进行判断，最重要的就是一直维护每个顶点的入度序列，每次将一个入度数为0的顶点弹出，就更新一次这个度序列
代码如下：
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;

}


总结：拓扑排序是深度优先的很好使用，针对不同的情况可以使用图的邻接表或者邻接矩阵表示，使得复杂度降低