到达第 N 个城市所需的最少桥梁

给定一个整数N表示连接城市的数量(从1到N编号)和一个二维数组arr[][]由通过双向桥相互连接的对组成。任务是找到从城市1到达城市N所需跨越的最少桥梁数量。

例子：

Input: N = 3, M = 2, arr[][] = {{1, 2}, {2, 3}}

Output: 2
Explanation: 
To reach Node 2 from Node 1, 1 bridge is required to be crossed. 
To reach Node 3 from Node 2, 1 bridge is required to be crossed.
Hence, 2 bridges are required to be connected.

Input: N = 4, M = 3, arr[][] = {{1, 2}, {2, 3}, {2, 4}}
Output: 2

处理方法：按照以下步骤解决问题：

• 初始化一个邻接表来构建和存储 Graph 节点。
• 初始化一个数组，比如大小为N 的vis[]来标记访问过的节点，另一个数组，比如大小为N 的dist[] ，用于存储到城市1的最小距离。
• 执行BFS并使用单源最短路径的概念来遍历图形并存储从城市1到达每个城市所需的最少桥梁数量。
• 打印dist[N]的值作为从城市1到城市N的最小距离。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Adjacency list to store graph
vector g[10001];
 
// Stores info about visited nodes
int vis[10001];
 
// Stores distance of nodes
// from the source node
int dist[10001];
 
// Function for BFS traversal
void BFS(int src)
{
    // Stores the nodes
    queue q;
 
    // Push the source node
    q.push(src);
 
    // Mark the pushed node visited
    vis[src] = 1;
 
    // Source node is always at dist 0
    dist[src] = 0;
 
    // Iterate until queue is not empty
    while (!q.empty()) {
 
        // Update the current node
        int curr = q.front();
 
        // Pop the node after
        // update by curr
        q.pop();
 
        // Traverse every node of
        // the adjacency list
        for (auto child : g[curr]) {
            if (vis[child] == 0) {
 
                // Push the child node
                // if its not visited
                q.push(child);
 
                // Update the distance of next level
                // nodes as it can be accessed by the
                // previous node in BFS
                dist[child] = dist[curr] + 1;
 
                // Mark the child node as visited
                vis[child] = 1;
            }
        }
    }
}
 
// Function to build the graph
void buildGraph(int M, int arr[][2])
{
    for (int i = 0; i < M; i++) {
        g[arr[i][0]].push_back(arr[i][1]);
        g[arr[i][1]].push_back(arr[i][0]);
    }
}
 
// Function to print the distance between from
// city 1 to city N
void shortestDistance(int N, int M, int arr[][2])
{
    // Build graph
    buildGraph(M, arr);
 
    // Perform BFS traversal
    BFS(1);
 
    // Print the shortest distance
    cout << dist[N];
}
 
// Driver Code
int main()
{
    // Given number of Nodes & Edges
    int N = 3, M = 2;
 
    // Given pairs of edges
    int arr[][2] = { { 1, 2 }, { 2, 3 } };
 
    // Function Call
    shortestDistance(N, M, arr);
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Adjacency list to store graph
static Vector []g = new Vector[10001];
 
// Stores info about visited nodes
static int []vis = new int[10001];
 
// Stores distance of nodes
// from the source node
static int []dist = new int[10001];
static {
    for(int i = 0; i < g.length; i++)
    {
        g[i] = new Vector<>();
    }
}
   
// Function for BFS traversal
static void BFS(int src)
{
   
    // Stores the nodes
    Queue q = new LinkedList<>();
 
    // Push the source node
    q.add(src);
 
    // Mark the pushed node visited
    vis[src] = 1;
 
    // Source node is always at dist 0
    dist[src] = 0;
 
    // Iterate until queue is not empty
    while (!q.isEmpty()) {
 
        // Update the current node
        int curr = q.peek();
 
        // Pop the node after
        // update by curr
        q.remove();
 
        // Traverse every node of
        // the adjacency list
        for (int child : g[curr]) {
            if (vis[child] == 0) {
 
                // Push the child node
                // if its not visited
                q.add(child);
 
                // Update the distance of next level
                // nodes as it can be accessed by the
                // previous node in BFS
                dist[child] = dist[curr] + 1;
 
                // Mark the child node as visited
                vis[child] = 1;
            }
        }
    }
}
 
// Function to build the graph
static void buildGraph(int M, int arr[][])
{
    for (int i = 0; i < M; i++) {
        g[arr[i][0]].add(arr[i][1]);
        g[arr[i][1]].add(arr[i][0]);
    }
}
 
// Function to print the distance between from
// city 1 to city N
static void shortestDistance(int N, int M, int arr[][])
{
   
    // Build graph
    buildGraph(M, arr);
 
    // Perform BFS traversal
    BFS(1);
 
    // Print the shortest distance
    System.out.print(dist[N]);
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given number of Nodes & Edges
    int N = 3, M = 2;
 
    // Given pairs of edges
    int arr[][] = { { 1, 2 }, { 2, 3 } };
 
    // Function Call
    shortestDistance(N, M, arr);
}
}
 
// This code is contributed by shikhasingrajput.

Python3

# Python 3 program for the above approach
 
# Adjacency list to store graph
g = [[] for i in range(10001)]
 
# Stores info about visited nodes
vis = [0 for i in range(10001)]
 
# Stores distance of nodes
# from the source node
dist = [0 for i in range(10001)]
 
# Function for BFS traversal
def BFS(src):
    global vis
    global dist
    global g
     
    # Stores the nodes
    q = []
 
    # Push the source node
    q.append(src)
 
    # Mark the pushed node visited
    vis[src] = 1
 
    # Source node is always at dist 0
    dist[src] = 0
 
    # Iterate until queue is not empty
    while (len(q)):
       
        # Update the current node
        curr = q[0]
 
        # Pop the node after
        # update by curr
        q.remove(q[0])
 
        # Traverse every node of
        # the adjacency list
        for child in g[curr]:
            if (vis[child] == 0):
               
                # Push the child node
                # if its not visited
                q.append(child)
 
                # Update the distance of next level
                # nodes as it can be accessed by the
                # previous node in BFS
                dist[child] = dist[curr] + 1
 
                # Mark the child node as visited
                vis[child] = 1
 
# Function to build the graph
def buildGraph(M, arr):
    global g
    for i in range(M):
        g[arr[i][0]].append(arr[i][1])
        g[arr[i][1]].append(arr[i][0])
 
# Function to print the distance between from
# city 1 to city N
def shortestDistance(N, M, arr):
   
    # Build graph
    buildGraph(M, arr)
 
    # Perform BFS traversal
    BFS(1)
     
    # Print the shortest distance
    print(dist[N])
 
# Driver Code
if __name__ == '__main__':
   
    # Given number of Nodes & Edges
    N = 3
    M = 2
 
    # Given pairs of edges
    arr =  [[1, 2], [2, 3]]
 
    # Function Call
    shortestDistance(N, M, arr)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
// Adjacency list to store graph
static List []g = new List[10001];
 
// Stores info about visited nodes
static int []vis = new int[10001];
 
// Stores distance of nodes
// from the source node
static int []dist = new int[10001];
 
   
// Function for BFS traversal
static void BFS(int src)
{
   
    // Stores the nodes
    Queue q = new Queue();
 
    // Push the source node
    q.Enqueue(src);
 
    // Mark the pushed node visited
    vis[src] = 1;
 
    // Source node is always at dist 0
    dist[src] = 0;
 
    // Iterate until queue is not empty
    while (q.Count!=0) {
 
        // Update the current node
        int curr = q.Peek();
 
        // Pop the node after
        // update by curr
        q.Dequeue();
 
        // Traverse every node of
        // the adjacency list
        foreach (int child in g[curr]) {
            if (vis[child] == 0) {
 
                // Push the child node
                // if its not visited
                q.Enqueue(child);
 
                // Update the distance of next level
                // nodes as it can be accessed by the
                // previous node in BFS
                dist[child] = dist[curr] + 1;
 
                // Mark the child node as visited
                vis[child] = 1;
            }
        }
    }
}
 
// Function to build the graph
static void buildGraph(int M, int [,]arr)
{
    for (int i = 0; i < M; i++) {
        g[arr[i,0]].Add(arr[i,1]);
        g[arr[i,1]].Add(arr[i,0]);
    }
}
 
// Function to print the distance between from
// city 1 to city N
static void shortestDistance(int N, int M, int [,]arr)
{
   
    // Build graph
    buildGraph(M, arr);
 
    // Perform BFS traversal
    BFS(1);
 
    // Print the shortest distance
    Console.Write(dist[N]);
}
 
// Driver Code
public static void Main(String[] args)
{
   
    // Given number of Nodes & Edges
    int N = 3, M = 2;
 
    // Given pairs of edges
    int [,]arr = { { 1, 2 }, { 2, 3 } };
 
    for(int i = 0; i < g.Length; i++)
    {
        g[i] = new List();
    }
    // Function Call
    shortestDistance(N, M, arr);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// JavaScript program for the above approach
 
// Adjacency list to store graph
var g = Array.from(Array(10001), ()=>new Array());;
 
// Stores info about visited nodes
var vis = Array(10001).fill(false);
 
// Stores distance of nodes
// from the source node
var dist = Array(10001).fill(0);
 
// Function for BFS traversal
function BFS(src)
{
    // Stores the nodes
    var q = [];
 
    // Push the source node
    q.push(src);
 
    // Mark the pushed node visited
    vis[src] = 1;
 
    // Source node is always at dist 0
    dist[src] = 0;
 
    // Iterate until queue is not empty
    while (q.length!=0) {
 
        // Update the current node
        var curr = q[0];
 
        // Pop the node after
        // update by curr
        q.shift();
 
        // Traverse every node of
        // the adjacency list
        g[curr].forEach(child => {
              if (vis[child] == 0) {
 
                // Push the child node
                // if its not visited
                q.push(child);
 
                // Update the distance of next level
                // nodes as it can be accessed by the
                // previous node in BFS
                dist[child] = dist[curr] + 1;
 
                // Mark the child node as visited
                vis[child] = 1;
            }
        });
     
    }
}
 
// Function to build the graph
function buildGraph(M, arr)
{
    for (var i = 0; i < M; i++) {
        g[arr[i][0]].push(arr[i][1]);
        g[arr[i][1]].push(arr[i][0]);
    }
}
 
// Function to print the distance between from
// city 1 to city N
function shortestDistance(N, M, arr)
{
    // Build graph
    buildGraph(M, arr);
 
    // Perform BFS traversal
    BFS(1);
 
    // Print the shortest distance
    document.write( dist[N]);
}
 
// Driver Code
// Given number of Nodes & Edges
var N = 3, M = 2;
 
// Given pairs of edges
var arr = [ [ 1, 2 ], [ 2, 3 ] ];
 
// Function Call
shortestDistance(N, M, arr);
 
 

输出：

2

时间复杂度： O(N)
辅助空间： O(N)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live