LeetCode 207. Course Schedule

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.

问题描述

给定n门课程，编号为0~n-1，然后某些课有其先修课程，求在给定n门课的先后关系中，能否完成所有的课程？

例如：输入2 表示两门课程，[[1,0]]表示，在修课程1之前，必须修过课程0。

解题思路

其实就是一道简单的拓扑排序问题。有几个点需要注意：
1、如何存储图的结构——本题解法用二维vector（连接表）
2、入度数组degree 统计
3、是否遍历过某个课程——标记数组（标记数组可以另开一个vis数组，但是本题中可以用degree数组标记为-1表示已遍历该课程）
解题思路：就是求入度为0 的课程，放入队列中，然后将与该课程关联的课程的入度-1，如果关联课程入度为0，放入队列
代码如下：

bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) {
queue<int> Q;
int degree[numCourses];
memset(degree,0,sizeof(degree));
vector<vector<int> > Graph(numCourses);
vector<pair<int, int> >::iterator iter;
// 首先存储图结构
for (iter = prerequisites.begin(); iter != prerequisites.end(); ++iter) {
++degree[iter->first];
Graph[iter->second].push_back(iter->first);
}
// 判断入度为0的点
for (int i = 0; i < numCourses; ++i) {
if (degree[i] == 0) {
Q.push(i);
degree[i] = -1;
}
}
// 拓扑
while(!Q.empty()) {
int x = Q.front();
Q.pop();
for(int i = 0; i < Graph[x].size(); ++i) {
--degree[Graph[x][i]];
if (degree[Graph[x][i]] == 0) {
Q.push(Graph[x][i]);
degree[Graph[x][i]] = -1;
}
}
}
// 判断是否遍历了所有点
for(int i = 0; i < numCourses; ++i) {
if(degree[i] != -1)
return false;
}
return true;
}

LeetCode 运行结果：



至此：本题解答完毕。
如果您有更优秀的想法，欢迎一起交流！