[leetcode] 207.Course Schedule

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?

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.

题意：

有n门课程，编号是0到n-1。有些课程需要依赖其他先修课程修完才能再去修读，比如对于一个pair[0,1]代表就是要修读课程1必须先修读课程0。

判断n门课程能否都修读完，比如[0,1],[1,0]，意思1号课程需要依赖0号课程，并且0号课程需要依赖1号。

思路：

这是一道关于拓扑排序的问题，我们可以使用深度优先或者广度优先遍历方法。先将不需要先修课程的课程号装入，然后将其出队列，并且将所有依赖这门课的课程的依赖数减去1，如果减之后变成了0，那么将这门课程加入队列。因为此时代表该课程的先修科目已经全部修完。

以上。

代码如下：

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
int size = prerequisites.size();
if(size == 0)return true;
vector<int> degree(numCourses, 0);
unordered_multimap<int,int> preMap;
for(auto ele:prerequisites)
{
degree[ele.second]++;
preMap.insert(ele);
}
int count = 0;
queue<int> que;
for(int i = 0; i < numCourses; i++)
{
if(degree[i] == 0)
que.push(i);
}
while(!que.empty())
{
count++;
int pre = que.front();
que.pop();
auto res = preMap.equal_range(pre);
for(auto beg = res.first; beg != res.second; beg++)
{
degree[beg->second]--;
if(degree[beg->second] == 0)
que.push(beg->second);
}
}
if(count!= numCourses)return false;
else return true;
}
};