广度优先搜索_钥匙和房间

841. 钥匙和房间 - 力扣（LeetCode）

算法描述

起始元素入队 查找队首元素的邻居节点 发现之后 入队 并且标记位已访问 

接着 队首元素出队 一直循环 直到队列为空

#define maxn 10001

class Solution {
public:

    void addEdge( vector<vector<int>>& table, vector<vector<int>>& rooms, int n){
        for ( int i = 0; i < rooms.size(); ++i){
            for ( int j : rooms[i])
            table[i].push_back(j);
        }
    }

    int bfs( vector<vector<int>>& table, int n, int visited[]){

        queue<int> q;
        q.push(0);
        visited[0] = true;
        int cnt = 0;
        while ( !q.empty() ){
            ++cnt;
            int u = q.front();
            q.pop();
            for ( int i = 0; i < table[u].size(); ++i ){
                int j = table[u][i];
                if ( !visited[j]){
                    q.push(j);
                    visited[j] = true;
                }
            }
        }
        return cnt; 
    }

    bool canVisitAllRooms(vector<vector<int>>& rooms) {
        int n = rooms.size();

        vector< vector<int> > table(n);
        int visited[maxn] = {0};

        addEdge( table, rooms, n);

        return bfs( table, n, visited) == n;


    }
};