leetcode_c++:图：Course Schedule II (207)

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

跟 207 一样拓扑排序，不过要记录路径。

用栈，只要每次记一下节点就行了。

用 dfs，不能跟 207 一样直接一个一个 check 过去。先找出入度为 0 的节点，再 dfs 过去。dfs 中每个节点处理完子节点后再把自己节点加到路径中，这样就维护了一个倒序的路径，最后把这个路径反向一下就行了。

#include <bits/stdc++.h>

using namespace std;
const int N = 0;

class Solution {
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
graph = vector<vector<int> >(numCourses);
vis = vector<int>(numCourses);

vector<int> inorder(numCourses);
vector<int> path;
// generate the tree map
for (auto prerequisite : prerequisites) {
graph[prerequisite.second].push_back(prerequisite.first);
inorder[prerequisite.first] += 1;
}
// find out 0-in-order node and dfs
for (int i = 0; i < numCourses; ++i) {
if (inorder[i] == 0) {
if (dfs(i, path) == false) // can not topsort
return vector<int>();
}
}
if (path.size() != numCourses)
return vector<int>();
reverse(path.begin(), path.end());
return path;
}
private:
vector<vector<int> > graph;
vector<int> vis;

bool dfs(int id, vector<int> &path) {
vis[id] = 1; // visiting
for (auto toid : graph[id]) {
if (vis[toid] == 1)
return false;
if (vis[toid] == 2)
continue;
if (dfs(toid, path) == false)
return false;
}
vis[id] = 2; // visited
path.push_back(id);
return true;
}
};

int main() {
Solution s;
vector<pair<int, int>> prerequisites;
prerequisites.push_back(pair<int, int>(0, 2));
prerequisites.push_back(pair<int, int>(2, 0));
prerequisites.push_back(pair<int, int>(1, 2));
vector<int> res = s.findOrder(3, prerequisites);
for (auto x : res) cout << x << ' ';
return 0;
}