Question

我有一个包含源 - 目标节点和阈值的大文本文件（20 GB）文件。如果阈值＆gt; 0连接，否则没有连接。我想在数组或数组列表中添加连接的节点（哪一个适合大量的数据？）并找到巨大的连接组件。我认为BFS算法是最短路径的解决方案

Txt文件

100 101 -0.3434
100 102  1.0023
100 103  1.100
103 104  0.210
...

我的代码：

String line = null;
HashMap<Integer,ArrayList<Node>> arr = new HashMap<Integer,ArrayList<Node>>();
BufferedReader reader = new BufferedReader(new FileReader("C:/Users/UserPC/Desktop/output.txt"));
        while((line = reader.readLine()) != null){
            String[] spl = line.split("\\s+");
            //System.out.println(spl[0]+","+spl[1]);
            int source = Integer.parseInt(spl[0]);
            int target = Integer.parseInt(spl[1]);


            arr.computeIfAbsent( source, k -> new ArrayList<>()).add(new Node(target));

        } 

        reader.close();

Answer 1

如果你没有正确实现它们，20 GB会花费太多时间，因为你可以考虑使用图形数据结构，然后通过调整权重来应用最小spanningtree算法，这样它只能找到权重小于或大于某个权重的节点值。

步骤：步骤1：制作包含源节点，目的地和重量/阈值的一维行李袋。

int firstNode = readIn via scanner.
int secondNode = readIn.
Int thresHold = read;
Connected connected = new Conncted(firstNode,secondNode,threshold);



    add all these connected component into array of bag so you have a graph then use minimumspanning tree or anyother algorithm, there are many.

创建一个类，表示您的组件已连接，其阈值是什么：呼叫已连接。

const spawnTime = 10000;

var coin = 0;
var intervalId = '';

var coinDiv = $('#coinDiv');
var coinImg = $('#coinImg');
var invDiv  = $('#invDiv');
var invId   = $('#inventoryId');
var invImg  = $('#invLocation');


coinImg.on('click', collect);
intervalId = setInterval(setLocation, spawnTime);


function setLocation() {
  var x = parseInt( Math.random()*(80-20+1) ) + 20;
  var y = parseInt( Math.random()*(80-20+1) ) + 20;

  coinImg.animate({
    opacity: 1
  }, 3000,
  function() {
    coinImg.css('top', x+'%');
    coinImg.css('left', y+'%');
    coinImg.css('display', 'initial');

    setTimeout( () => coinImg.animate({ opacity: 0 }, 3000), 6000);
  });
}

function collect() {
    clearInterval(intervalId);

  coinImg.stop();
  coinImg.css('opacity', 1);

  /* Increment coin counter */
  coin++;
  invId.text(coin);

  /* In order to disable multiple clicks */
  coinImg.css('pointer-events', 'none');

  /* Double the size */
  coinImg.css('width', '128px');
  coinImg.css('height', '128px');

  /* Animate and return to normal */
  coinImg.animate({
        width:  '32px',
        height: '32px',
        left:   invImg.offset().left + 'px',
        top:    invImg.offset().top  + 'px'
    }, 1500,
    function() {
      coinImg.css('pointer-events', 'auto');
      coinImg.css('display', 'none');
      coinImg.css('width', '64px');
      coinImg.css('height', '64px');

      intervalId = setInterval(setLocation, spawnTime);
    }
  );
}

网络中最大的连接组件

1 个答案: