给定一个无向图，任务是计算给定图中任意两个顶点之间的桥的最大数量。
例子：
Input:
Graph
1 ------- 2 ------- 3 -------- 4
| |
| |
5 ------- 6
Output: 2
Explanation:
There are 2 bridges, (1 - 2)
and (3 - 4), in the path from 1 to 4.
Input:
Graph:
1 ------- 2 ------- 3 ------- 4
Output: 3
Explanation:
There are 3 bridges, (1 - 2), (2 - 3)
and (3 - 4) in the path from 1 to 4.
方法：
请按照以下步骤解决问题：
- 找到图中的所有桥并将它们存储在一个向量中。
- 移除所有桥接将图形缩减为小组件。
- 这些小组件没有任何桥接，它们是不包含桥接的弱连接组件。
- 生成由桥连接的节点组成的树，以桥为边。
- 现在，任何节点之间的路径中的最大桥梁等于这棵树的直径。
- 因此，找到这棵树的直径并将其打印为答案。
下面是上述方法的实现
C++
// C++ program to find the
// maximum number of bridges
// in any path of the given graph
#include
using namespace std;
const int N = 1e5 + 5;
// Stores the nodes
// and their connections
vector > v(N);
// Store the tree with
// Bridges as the edges
vector > g(N);
// Stores the visited nodes
vector vis(N, 0);
// for finding bridges
vector in(N), low(N);
// for Disjoint Set Union
vector parent(N), rnk(N);
// for storing actaul bridges
vector > bridges;
// Stores the number of
// nodes and edges
int n, m;
// For finding bridges
int timer = 0;
int find_set(int a)
{
// Function to find root of
// the component in which
// A lies
if (parent[a] == a)
return a;
// Doing path compression
return parent[a]
= find_set(parent[a]);
}
void union_set(int a, int b)
{
// Function to do union
// between a and b
int x = find_set(a), y = find_set(b);
// If both are already in the
// same component
if (x == y)
return;
// If both have same rank,
// then increase anyone's rank
if (rnk[x] == rnk[y])
rnk[x]++;
if (rnk[y] > rnk[x])
swap(x, y);
parent[y] = x;
}
// Function to find bridges
void dfsBridges(int a, int par)
{
vis[a] = 1;
// Initialize in time and
// low value
in[a] = low[a] = timer++;
for (int i v[a]) {
if (i == par)
continue;
if (vis[i])
// Update the low value
// of the parent
low[a] = min(low[a], in[i]);
else {
// Perform DFS on its child
// updating low if the child
// has connection with any
// ancestor
dfsBridges(i, a);
low[a] = min(low[a], low[i]);
if (in[a] < low[i])
// Bridge found
bridges.push_back(make_pair(i, a));
// Otherwise
else
// Find union between parent
// and child as they
// are in same component
union_set(i, a);
}
}
}
// Function to find diameter of the
// tree for storing max two depth child
int dfsDiameter(int a, int par, int& diameter)
{
int x = 0, y = 0;
for (int i g[a]) {
if (i == par)
continue;
int mx = dfsDiameter(i, a, diameter);
// Finding max two depth
// from its children
if (mx > x) {
y = x;
x = mx;
}
else if (mx > y)
y = mx;
}
// Update diameter with the
// sum of max two depths
diameter = max(diameter, x + y);
// Return the maximum depth
return x + 1;
}
// Function to find maximum
// bridges bwtween
// any two nodes
int findMaxBridges()
{
for (int i = 0; i <= n; i++) {
parent[i] = i;
rnk[i] = 1;
}
// DFS to find bridges
dfsBridges(1, 0);
// If no bridges are found
if (bridges.empty())
return 0;
int head = -1;
// Iterate over all bridges
for (auto& i bridges) {
// Find the endpoints
int a = find_set(i.first);
int b = find_set(i.second);
// Generate the tree with
// bridges as the edges
g[a].push_back(b);
g[b].push_back(a);
// Update the head
head = a;
}
int diameter = 0;
dfsDiameter(head, 0, diameter);
// Return the diameter
return diameter;
}
// Driver Code
int main()
{
/*
Graph =>
1 ---- 2 ---- 3 ---- 4
| |
5 ---- 6
*/
n = 6, m = 6;
v[1].push_back(2);
v[2].push_back(1);
v[2].push_back(3);
v[3].push_back(2);
v[2].push_back(5);
v[5].push_back(2);
v[5].push_back(6);
v[6].push_back(5);
v[6].push_back(3);
v[3].push_back(6);
v[3].push_back(4);
v[4].push_back(4);
int ans = findMaxBridges();
cout << ans << endl;
return 0;
} Java
// Java program to find the
// maximum number of bridges
// in any path of the given graph
import java.util.*;
class GFG{
static int N = (int)1e5 + 5;
// Stores the nodes
// and their connections
static Vector []v =
new Vector[N];
// Store the tree with
// Bridges as the edges
static Vector []g =
new Vector[N];
// Stores the visited nodes
static boolean []vis =
new boolean[N];
// For finding bridges
static int []in = new int[N];
static int []low = new int[N];
// for Disjoint Set Union
static int []parent = new int[N];
static int []rnk = new int[N];
// For storing actaul bridges
static Vector bridges =
new Vector<>();
// Stores the number of
// nodes and edges
static int n, m;
// For finding bridges
static int timer = 0;
static int diameter;
static class pair
{
int first, second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
static void swap(int x,
int y)
{
int temp = x;
x = y;
y = temp;
}
static int find_set(int a)
{
// Function to find root of
// the component in which
// A lies
if (parent[a] == a)
return a;
// Doing path compression
return parent[a] =
find_set(parent[a]);
}
static void union_set(int a, int b)
{
// Function to do union
// between a and b
int x = find_set(a),
y = find_set(b);
// If both are already
// in the same component
if (x == y)
return;
// If both have same rank,
// then increase anyone's rank
if (rnk[x] == rnk[y])
rnk[x]++;
if (rnk[y] > rnk[x])
swap(x, y);
parent[y] = x;
}
// Function to find bridges
static void dfsBridges(int a,
int par)
{
vis[a] = true;
// Initialize in time and
// low value
in[a] = low[a] = timer++;
for (int i : v[a])
{
if (i == par)
continue;
if (vis[i])
// Update the low value
// of the parent
low[a] = Math.min(low[a],
in[i]);
else
{
// Perform DFS on its child
// updating low if the child
// has connection with any
// ancestor
dfsBridges(i, a);
low[a] = Math.min(low[a],
low[i]);
if (in[a] < low[i])
// Bridge found
bridges.add(new pair(i, a));
// Otherwise
else
// Find union between parent
// and child as they
// are in same component
union_set(i, a);
}
}
}
// Function to find diameter
// of the tree for storing
// max two depth child
static int dfsDiameter(int a,
int par)
{
int x = 0, y = 0;
for (int i : g[a])
{
if (i == par)
continue;
int mx = dfsDiameter(i, a);
// Finding max two depth
// from its children
if (mx > x)
{
y = x;
x = mx;
}
else if (mx > y)
y = mx;
}
// Update diameter with the
// sum of max two depths
diameter = Math.max(diameter,
x + y);
// Return the maximum depth
return x + 1;
}
// Function to find maximum
// bridges bwtween
// any two nodes
static int findMaxBridges()
{
for (int i = 0; i <= n; i++)
{
parent[i] = i;
rnk[i] = 1;
}
// DFS to find bridges
dfsBridges(1, 0);
// If no bridges are found
if (bridges.isEmpty())
return 0;
int head = -1;
// Iterate over all bridges
for (pair i : bridges)
{
// Find the endpoints
int a = find_set(i.first);
int b = find_set(i.second);
// Generate the tree with
// bridges as the edges
g[a].add(b);
g[b].add(a);
// Update the head
head = a;
}
diameter = 0;
dfsDiameter(head, 0);
// Return the diameter
return diameter;
}
// Driver Code
public static void main(String[] args)
{
/*
Graph =>
1 ---- 2 ---- 3 ---- 4
| |
5 ---- 6
*/
n = 6;
m = 6;
for (int i = 0; i < v.length; i++)
v[i] = new Vector();
for (int i = 0; i < g.length; i++)
g[i] = new Vector();
v[1].add(2);
v[2].add(1);
v[2].add(3);
v[3].add(2);
v[2].add(5);
v[5].add(2);
v[5].add(6);
v[6].add(5);
v[6].add(3);
v[3].add(6);
v[3].add(4);
v[4].add(4);
int ans = findMaxBridges();
System.out.print(ans + "\n");
}
}
// This code is contributed by Princi Singh C#
// C# program to find the
// maximum number of bridges
// in any path of the given graph
using System;
using System.Collections.Generic;
class GFG{
static int N = (int)1e5 + 5;
// Stores the nodes
// and their connections
static List []v =
new List[N];
// Store the tree with
// Bridges as the edges
static List []g =
new List[N];
// Stores the visited nodes
static bool []vis =
new bool[N];
// For finding bridges
static int []init = new int[N];
static int []low = new int[N];
// for Disjoint Set Union
static int []parent = new int[N];
static int []rnk = new int[N];
// For storing actaul bridges
static List bridges =
new List();
// Stores the number of
// nodes and edges
static int n, m;
// For finding bridges
static int timer = 0;
static int diameter;
class pair
{
public int first, second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
static void swap(int x,
int y)
{
int temp = x;
x = y;
y = temp;
}
static int find_set(int a)
{
// Function to find root of
// the component in which
// A lies
if (parent[a] == a)
return a;
// Doing path compression
return parent[a] =
find_set(parent[a]);
}
static void union_set(int a,
int b)
{
// Function to do union
// between a and b
int x = find_set(a),
y = find_set(b);
// If both are already
// in the same component
if (x == y)
return;
// If both have same rank,
// then increase anyone's rank
if (rnk[x] == rnk[y])
rnk[x]++;
if (rnk[y] > rnk[x])
swap(x, y);
parent[y] = x;
}
// Function to find bridges
static void dfsBridges(int a,
int par)
{
vis[a] = true;
// Initialize in time and
// low value
init[a] = low[a] = timer++;
foreach (int i in v[a])
{
if (i == par)
continue;
if (vis[i])
// Update the low value
// of the parent
low[a] = Math.Min(low[a],
init[i]);
else
{
// Perform DFS on its child
// updating low if the child
// has connection with any
// ancestor
dfsBridges(i, a);
low[a] = Math.Min(low[a],
low[i]);
if (init[a] < low[i])
// Bridge found
bridges.Add(new pair(i, a));
// Otherwise
else
// Find union between parent
// and child as they
// are in same component
union_set(i, a);
}
}
}
// Function to find diameter
// of the tree for storing
// max two depth child
static int dfsDiameter(int a,
int par)
{
int x = 0, y = 0;
foreach (int i in g[a])
{
if (i == par)
continue;
int mx = dfsDiameter(i, a);
// Finding max two depth
// from its children
if (mx > x)
{
y = x;
x = mx;
}
else if (mx > y)
y = mx;
}
// Update diameter with the
// sum of max two depths
diameter = Math.Max(diameter,
x + y);
// Return the
// maximum depth
return x + 1;
}
// Function to find maximum
// bridges bwtween
// any two nodes
static int findMaxBridges()
{
for (int i = 0; i <= n; i++)
{
parent[i] = i;
rnk[i] = 1;
}
// DFS to find bridges
dfsBridges(1, 0);
// If no bridges are found
if (bridges.Count == 0)
return 0;
int head = -1;
// Iterate over all bridges
foreach (pair i in bridges)
{
// Find the endpoints
int a = find_set(i.first);
int b = find_set(i.second);
// Generate the tree with
// bridges as the edges
g[a].Add(b);
g[b].Add(a);
// Update the head
head = a;
}
diameter = 0;
dfsDiameter(head, 0);
// Return the diameter
return diameter;
}
// Driver Code
public static void Main(String[] args)
{
/*
Graph =>
1 ---- 2 ---- 3 ---- 4
| |
5 ---- 6
*/
n = 6;
m = 6;
for (int i = 0; i < v.Length; i++)
v[i] = new List();
for (int i = 0; i < g.Length; i++)
g[i] = new List();
v[1].Add(2);
v[2].Add(1);
v[2].Add(3);
v[3].Add(2);
v[2].Add(5);
v[5].Add(2);
v[5].Add(6);
v[6].Add(5);
v[6].Add(3);
v[3].Add(6);
v[3].Add(4);
v[4].Add(4);
int ans = findMaxBridges();
Console.Write(ans + "\n");
}
}
// This code is contributed by Amit Katiyar 输出：
2
时间复杂度： O(N + M)
辅助空间： O(N + M)
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