如何仅在两个节点之间保留边缘？

Question

我有追踪图。在我的示例中，我只需要保留位于两个节点Company 5和Company 7之间的年龄。喜欢：

我找不到任何示例该怎么做的问题。有人可以帮我吗？

实时示例：https://jsfiddle.net/5ntzqafv/

// create an array with nodes
var nodes = new vis.DataSet([
                {id: 1, label: 'Company 1', group: "company", },
                {id: 2, label: 'Mike', group: "owner"},
                {id: 3, label: 'David', group: "founder"},
                {id: 4, label: 'Company 2', group: "company"},
                {id: 5, label: 'Company 3', group: "company"},
                {id: 6, label: 'Company 4', group: "company"},
                {id: 8, label: 'Company 5', group: "company", borderWidth: 4, color: { border: '#077eb8' }, font: { size: 16},},
                {id: 9, label: 'Company 6', group: "company"},
                {id: 10, label: 'Company 7', group: "company", borderWidth: 4, color: { border: '#077eb8' }, font: { size: 16},},
                {id: 11, label: 'Company 8', group: "company"},
          {id: 12, label: 'Company 9', group: "company"}
]);

// create an array with edges
var edges = new vis.DataSet([
                {from: 1, to: 2},
                {from: 1, to: 3},
                {from: 2, to: 11},
                {from: 3, to: 4},
          {from: 4, to: 3},
                {from: 2, to: 8}, 
                {from: 3, to: 6},
                {from: 3, to: 5},
                {from: 4, to: 9},
          {from: 7, to: 4},
          {from: 7, to: 12},
          {from: 10, to: 12},
          {from: 9, to: 12},
                {from: 4, to: 10}
]);

// create a network
var container = document.getElementById("mynetwork");
var data = {
  nodes: nodes,
  edges: edges
};
var options = {nodes: {
        shape: "box",
        widthConstraint: {
          maximum: 200
        }
      }};
var network = new vis.Network(container, data, options);

Answer 1

您可以使用一些简单的算法来搜索图形中的路径，例如BFS。

此处描述了算法：https://en.wikipedia.org/wiki/Breadth-first_search

我的实现：https://jsfiddle.net/alexey_kuldoshin/3svmqujz/74/

function getPath(graph, from, to) {
    let queue = [ from ];
    let p = {};
    p[from] = -1;
    while (queue.length > 0) {
        let v = queue.shift();
        if (v == to) {
            break;
        }
        graph[v].forEach(edge => {
            if (!(edge in p)) {
                p[edge] = v;
                queue.push(edge);
            }
        });
    }
    let ans = [];
    while (to != -1) {
        ans.push(to);
        to = p[to];
    }
    console.log(ans);
    return ans;
}