LeetCode -- 207. 课程表 （拓扑排序）

 经典拓扑排序的应用，用拓扑排序的算法看看原图中是否有一个合法的拓扑序。

class Solution {
public:
    const static int N = 2010, M = 5010;
    int h[N], e[M], ne[M], idx;
    int d[N], q[N];

    void add(int a, int b) {
        e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
    }

    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        memset(h, -1, sizeof h);

        for(auto &it : prerequisites) {
            add(it[1], it[0]);
            d[it[0]] ++ ;
        }

        auto f = [&]() {
            int hh = 0, tt = -1;
            for(int i = 0; i < numCourses; i ++ ) {
                if(!d[i]) q[ ++ tt] = i;
            }

            while(hh <= tt) {
                int t = q[hh ++ ];
                for(int i = h[t]; ~i; i = ne[i]) {
                    int j = e[i];
                    if( -- d[j] == 0) q[ ++ tt] = j;
                }
            }

            return tt == numCourses - 1;
        };

        return f();

    }
};

 

class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
        din = [0] * numCourses
        queue = deque()
        edges = [[] for _ in range(numCourses)]
        for cur, pre in prerequisites:
            din[cur] += 1
            edges[pre].append(cur)
        
        cnt = 0
        for i in range(len(din)):
            if not din[i]:
                queue.append(i)
                cnt += 1

        while queue:
            t = queue.popleft()
            for cur in edges[t]:
                din[cur] -= 1
                if not din[cur]:
                    queue.append(cur)
                    cnt += 1

        return cnt == numCourses