백준 5567번 결혼식 JavaScript 풀이 [BFS]

백준 5567번 결혼식 문제를 자바스크립트로 BFS 알고리즘을 이용해서 풀어보았다. 

https://www.acmicpc.net/problem/5567

 

let fs = require("fs");
let input = fs.readFileSync('/dev/stdin').toString().trim().split("\n");

class Queue {
  constructor() {
    this.items = {};
    this.head = 0;
    this.tail = 0;
  }
  enqueue(element) {
    this.items[this.tail] = element;
    this.tail++;
  }
  dequeue() {
    const element = this.items[this.head];
    delete this.items[this.head];
    this.head++;
    return element;
  }
  getLength() {
    return this.tail - this.head;
  }
}

let n = Number(input[0]);
let m = Number(input[1]);
let graph = new Array(n+1).fill().map(_ => []);
for (t=2; t<=m+1; t++) {
  let [f1, f2] = input[t].split(' ').map(Number);
  graph[f1].push(f2)
  graph[f2].push(f1)
}

queue = new Queue();
queue.enqueue(1);

let visited = new Array(n+1).fill(0);

while (queue.getLength() != 0) {
  let cur = queue.dequeue();

  for (next of graph[cur]) {
    if (visited[next] == 0) {
      visited[next] = visited[cur] + 1;
      queue.enqueue(next)
    }
  }
}

// 자기자신은 0으로 바꿔놓기
visited[1] = 0;
console.log(visited.filter(v => 1 <= v && v<=2).length)