LeetCode#207 Course Schedule

[Description]:

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.

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how
a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

[题解]:

给出一组上课的顺序，要求检查是否能依次上课。这就是拓扑排序。

[Solution]:
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<vector<int>> adj;
adj.resize(numCourses);
for (int i = 0; i < numCourses; i++) {
adj[i].resize(numCourses);
}
std::vector<int> in;
in.resize(numCourses);
for (int i = 0; i < prerequisites.size(); i++) {
int a = prerequisites[i].first; int b = prerequisites[i].second;
adj[a][b] = 1;
in[b]++;
}
std::queue<int> q;
for (int i = 0; i < numCourses; i++) {
if (in[i] == 0) q.push(i);
}
while (!q.empty()) {
int a = q.front();
q.pop();
for (int i = 0; i < numCourses; i++) {
if (adj[a][i] == 1) {
in[i]--;
if (in[i] == 0) {
q.push(i);
}
}
}
}

for (int i = 0; i < numCourses; i++) {
if (in[i] != 0) return false;
}
return true;
}
};