LeetCode Course Schedule BFS

思路：

BFS + 拓扑排序

Status：MLE
Reason：用二维数组 vector<vector<int>> 存图信息，有太多没用的信息0，占用了很大空间。
看了DISCUSS后的优化：
用 vector<unordered_set<int>> 存图信息，表示为每个节点对应连接到哪些节点。

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> inDegree(numCourses, 0);
        vector<vector<int>> graph(numCourses, vector<int>(numCourses, 0));
        queue<int> q;
        for(int i = 0; i < numCourses; ++i) {
            if(inDegree[i] == 0) q.push(i);
        }
        while(!q.empty()) {
            int start_node_id = q.front();
            q.pop();
            for(int i = 0; i < numCourses; ++i) {
                if(graph[start_node_id][i] == 1) {
                    inDegree[i]--;
                    if(inDegree[i] == 0) {
                        q.push(i);
                    }
                }
            }
        }
        for(int i = 0; i < numCourses; ++i) {
            if(inDegree[i] != 0) {
                return false;
            }
        }
        return true;
    }
};

Staus：AC

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> need(numCourses);
        for (size_t i = 0; i != prerequisites.size(); ++i)
            need[prerequisites[i].second].insert(prerequisites[i].first);
        vector<int> indegree(numCourses);
        for (int i = 0; i != numCourses; ++i)
            for(auto it = need[i].begin(); it != need[i].end(); ++it)
                ++indegree[*it];
        queue<int> zeros;
        for (int i = 0; i != numCourses; ++i)
            if (indegree[i] == 0)
                zeros.push(i);
        while (!zeros.empty()) {
            int seq = zeros.front();
            zeros.pop();
            for (auto it = need[seq].begin(); it != need[seq].end(); ++it)
                if (--indegree[*it] == 0)
                    zeros.push(*it);
            --numCourses;
        }
        return numCourses == 0;
    }
};