如何使用邻接矩阵找到未指向，未加权图的最重的路径（图的M（d））的重量

Question

图像中最长的路径是最长的路径，但LM是最重的 该图的最重的路径是连接到其最边缘的路径，对于此当前情况，最重的路径是LM，具有19个边缘：LK，KJ，JI，IH，IH，HG，HG，GF，GF，FE，FM，FM，FM，MN，MO，MO，MO，MO，MO，MO， MP，MQ，MR，MS，MT，MU，MV，MV，MW，MX。 我正在尝试创建一个程序，该程序将检查输入的矩阵是否代表树图，如果确实如此，则返回图形最重的路径上的边数。换句话说，它将返回图的m（d）。 对于当前情况，输出将是连接到LM路径的边数，即19岁。

但是我找不到这样做的方法。在这里，我创建了一个函数，以找到至少1秒以将其设置为根，但不知道该如何进行。

let matrix = [
  [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],


//find row with least 1s
function findRowWithLeastOnes(matrix_data) {
  let counter = 0;
  let row = 0;
  for (let i = 0; i < matrix_data.length; i++) {
    for (let j = 0; j < matrix_data[i].length; j++) {
      if (matrix_data[i][j] === 1) {
        counter++;
      }
    }
    if (counter < matrix_data.length) {
      row = i;
      console.log(row);
      return row;
    }
  }
}
findRowWithLeastOnes(matrix);

Answer 1

我们可以使用以下算法：

• 将矩阵变成邻接列表，以避免始终扫描零，并可以轻松访问节点的度，从而有助于重量。

• 选择以（例如0）开始的任何节点，然后从中找到最重的路径。我们可以对此进行深度优先搜索（例如，使用递归）。从发现的所有路径中，保持最重。为避免边缘来回穿越，请记住以前的节点在遍历中是什么：不应重新审视它。如果有一个先前的节点，我们应该避免在权重总和中对遍历边缘进行两次计数。

• 以该路径中的最后一个节点：这是远离我们开始节点的节点。该节点将是我们要寻找的最重的路径的终点。

• 第二次使用相同的算法来找到该节点最重的路径。这是我们需要的路径。

如果您的问题还涉及确定图是否是树，那么您本质上需要检查没有周期（深度第一个遍历可以再次使用），并且可以从一个节点中达到所有其他节点（首先是深度首先可以服务深度）。也许您还想检查图表是否没有方向，即矩阵沿主角镜像镜像。

function isUndirected(matrix) {
    if (matrix.length !== matrix[0]?.length) return false; // must be square
    for (let i = 0; i < matrix.length; i++) {
        for (let j = i; j < matrix.length; j++) {
            if (matrix[i][j] !== matrix[j][i]) return false;
        }
    }
    return true;
}

function isTree(matrix) {
    const visited = Array(matrix.length).fill(false);

    function dfs(node, previous=-1) {
        if (visited[node]) return false; // cycle
        visited[node] = true;
        const neighbors = matrix[node];
        for (let neighbor = 0; neighbor < neighbors.length; neighbor++) {
            if (neighbor != previous && neighbors[neighbor] && !dfs(neighbor, node)) return false;
        }
        return true;
    }
    // When no cycles were found and all nodes were reachable from node 0 => tree!
    return dfs(0) && !visited.includes(false);
}

function findHeaviestPathFrom(adj, node, previous=-1) {
    const neighbors = adj[node];
    let heaviest = {
        nodes: [],
        weight: 0
    };
    for (let neighbor of neighbors) {
        if (neighbor != previous) {
            const path = findHeaviestPathFrom(adj, neighbor, node);
            if (path.weight > heaviest.weight) heaviest = path;
        }
    }
    heaviest.nodes.push(node);
    heaviest.weight += neighbors.length - (previous !== -1)    
    return heaviest;
}

function findHeaviestPath(matrix) {
    // Turn the matrix into an adjacency list
    let adj = matrix.map(row => row.map((isEdge, i) => isEdge ? i : -1).filter(i => i > -1));
    // Find the node that has the heaviest connection to node 0
    let remote = findHeaviestPathFrom(adj, 0).nodes[0];
    // Find heaviest path starting in that node
    return findHeaviestPathFrom(adj, remote);
}

const matrix = [
  [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
];

if (!isUndirected(matrix)) console.log("The graph is directed");
else if (!isTree(matrix)) console.log("The graph is not a tree");
else console.log("Heaviest path: ", JSON.stringify(findHeaviestPath(matrix)));