给定两个N位数的素数A和B ，任务是找到将 A 转换为 B 所采取的最少步骤数。转换的条件是当前素数中只有 1 位可以修改，使得新数形成也是一个质数。如果没有这样的转换是可能的，打印 -1。
注： N 的取值范围为 [1, 5]。
例子：
Input: N = 4, A = 1033, B = 8179
Output: 6
Explanation: The steps of conversion are 1033 -> 1733 -> 3733 -> 3739 -> 3779 -> 8779 -> 8179. While changing numbers from 1033->1733, they differ by only one digit and the same happens for subsequent steps.
Input: N = 4, A = 1373, B = 8179
Output: 7
Input: N = 2, A = 11, B = 37
Output: 2
Explanation: The steps of conversion are 11 -> 17 -> 37
方法：使用广度优先搜索算法
- 找出所有 N 位素数，并用这些数作图。
- 将每个素数视为图的一个节点，如果它们相差一位数，则创建从一个节点到另一个节点的边。
- 应用 BFS 遍历并找出 A 和 B 之间的边数。
- 如果不存在路径，则打印 -1。
- 否则打印编号。的边缘，这是所需的解决方案。
下面是上述方法的实现。
C++
// C++ program for the above problem
#include
using namespace std;
#define ll long long
#define mod 1000000007
#define pb push_back
#define mod 1000000007
#define vi vector
// adjacency list for numbers
// till 100001
vi lis[100001];
vi primes;
// visited array
int vis[100001];
// to store distance of every
// vertex from A
int dis[100001];
// function to check if number
// is a prime
bool isPrime(int n)
{
for (int i = 2;
i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}
// function to check if numbers
// differ by only a single-digit
bool valid(int a, int b)
{
int c = 0;
while (a) {
// check the last digit of
// both numbers and increase
// count if different
if ((a % 10) != (b % 10)) {
c++;
}
a = a / 10;
b = b / 10;
}
if (c == 1) {
return true;
}
else {
return false;
}
}
void makePrimes(int N)
{
int i, j;
int L = pow(10, N - 1);
int R = pow(10, N) - 1;
// generate all N digit primes
for (int i = L; i <= R; i++) {
if (isPrime(i)) {
primes.pb(i);
}
}
for (i = 0;
i < primes.size(); i++) {
// for every prime number i
// check if an edge
// can be made.
for (j = i + 1;
j < primes.size(); j++) {
int a = primes[i];
int b = primes[j];
// for every prime number
// i check if an edge can
// be made from i to j.
if (valid(a, b)) {
// if edge is possible then
// insert in the adjacency
// list
lis[a].pb(b);
lis[b].pb(a);
}
}
}
}
// function to count distance
void bfs(int src)
{
queue q;
q.push(src);
vis[src] = 1;
dis[src] = 0;
while (!q.empty()) {
int curr = q.front();
q.pop();
for (int x : lis[curr]) {
// if unvisited push onto queue
// and mark visited as 1 and
// add the distance of curr+1.
if (vis[x] == 0) {
vis[x] = 1;
q.push(x);
dis[x] = dis[curr] + 1;
}
}
}
}
// Driver code
int main()
{
int N = 4;
makePrimes(N);
int A = 1033, B = 8179;
// Call bfs traversal
// with root as node A
bfs(A);
if (dis[B] == -1)
// Indicates not possible
cout << "-1" << endl;
else
cout << dis[B] << endl;
return 0;
} Java
// Java program for the above problem
import java.util.*;
public class GFG
{
// Adjacency list for numbers
// till 100001
static Vector> lis = new Vector>();
static Vector primes = new Vector();
// Visited array
static int[] vis = new int[100001];
// To store distance of every
// vertex from A
static int[] dis = new int[100001];
// Function to check if number
// is a prime
static boolean isPrime(int n)
{
int i = 2;
while (i * i <= n)
{
if (n % i == 0)
return false;
i += 1;
}
return true;
}
// Function to check if numbers
// differ by only a single-digit
static boolean valid(int a, int b)
{
int c = 0;
while(a > 0)
{
// Check the last digit of
// both numbers and increase
// count if different
if ((a % 10) != (b % 10))
c += 1;
a = a / 10;
b = b / 10;
}
if (c == 1)
return true;
else
return false;
}
static void makePrimes(int N)
{
// Generate all N digit primes
int L = (int)Math.pow(10, N - 1);
int R = (int)Math.pow(10, N) - 1;
for(int i = L; i < R + 1; i++)
{
if (isPrime(i))
primes.add(i);
}
for(int i = 0; i < primes.size(); i++)
{
// For every prime number i
// check if an edge
// can be made.
for(int j = i + 1; j < primes.size(); j++)
{
int a = primes.get(i);
int b = primes.get(j);
// For every prime number
// i check if an edge can
// be made from i to j.
if (valid(a, b))
{
// If edge is possible then
// insert in the adjacency
// list
lis.get(a).add(b);
lis.get(b).add(a);
}
}
}
}
// Function to count distance
static void bfs(int src)
{
Vector q = new Vector();
q.add(src);
vis[src] = 1;
dis[src] = 0;
while (q.size() != 0)
{
int curr = q.get(0);
q.remove(0);
for(int x : lis.get(curr))
{
// If unvisited push onto queue
// and mark visited as 1 and
// add the distance of curr+1.
if (vis[x] == 0)
{
vis[x] = 1;
q.add(x);
dis[x] = dis[curr] + 1;
}
}
}
}
// Driver code
public static void main(String[] args)
{
for(int i = 0; i < 100001; i++)
{
lis.add(new Vector());
}
int N = 4;
makePrimes(N);
int A = 1033;
int B = 8179;
// Call bfs traversal
// with root as node A
bfs(A);
if (dis[B] == -1)
{
// Indicates not possible
System.out.print(-1);
}
else
{
System.out.print(dis[B]);
}
}
}
// This code is contributed by divyesh072019 Python3
# Python3 program for the above problem
mod = 1000000007
# Adjacency list for numbers
# till 100001
lis = [[] for i in range(100001)]
primes = []
# Visited array
vis = [0 for i in range(100001)]
# To store distance of every
# vertex from A
dis = [0 for i in range(100001)]
# Function to check if number
# is a prime
def isPrime(n):
i = 2
while (i * i <= n):
if (n % i == 0):
return False
i += 1
return True
# Function to check if numbers
# differ by only a single-digit
def valid(a, b):
c = 0
while(a):
# Check the last digit of
# both numbers and increase
# count if different
if ((a % 10) != (b % 10)):
c += 1
a = int(a / 10)
b = int(b / 10)
if (c == 1):
return True
else:
return False
def makePrimes(N):
global primes
global lis
i = 0
j = 0
# Generate all N digit primes
L = pow(10, N - 1)
R = pow(10, N) - 1
for i in range(L, R + 1):
if (isPrime(i)):
primes.append(i)
for i in range(len(primes)):
# For every prime number i
# check if an edge
# can be made.
for j in range(i + 1, len(primes)):
a = primes[i]
b = primes[j]
# For every prime number
# i check if an edge can
# be made from i to j.
if (valid(a, b)):
# If edge is possible then
# insert in the adjacency
# list
lis[a].append(b)
lis[b].append(a)
# Function to count distance
def bfs(src):
global vis
global dis
q = []
q.append(src)
vis[src] = 1
dis[src] = 0
while (len(q) != 0):
curr = q[0]
q.pop(0)
for x in lis[curr]:
# If unvisited push onto queue
# and mark visited as 1 and
# add the distance of curr+1.
if (vis[x] == 0):
vis[x] = 1
q.append(x)
dis[x] = dis[curr] + 1
# Driver code
N = 4
makePrimes(N)
A = 1033
B = 8179
# Call bfs traversal
# with root as node A
bfs(A)
if (dis[B] == -1):
# Indicates not possible
print(-1)
else:
print(dis[B])
# This code is contributed by avanitrachhadiya2155C#
// C# program for the above problem
using System;
using System.Collections.Generic;
class GFG {
// Adjacency list for numbers
// till 100001
static List> lis = new List>();
static List primes = new List();
// Visited array
static int[] vis = new int[100001];
// To store distance of every
// vertex from A
static int[] dis = new int[100001];
// Function to check if number
// is a prime
static bool isPrime(int n)
{
int i = 2;
while (i * i <= n)
{
if (n % i == 0)
return false;
i += 1;
}
return true;
}
// Function to check if numbers
// differ by only a single-digit
static bool valid(int a, int b)
{
int c = 0;
while(a > 0)
{
// Check the last digit of
// both numbers and increase
// count if different
if ((a % 10) != (b % 10))
c += 1;
a = a / 10;
b = b / 10;
}
if (c == 1)
return true;
else
return false;
}
static void makePrimes(int N)
{
// Generate all N digit primes
int L = (int)Math.Pow(10, N - 1);
int R = (int)Math.Pow(10, N) - 1;
for(int i = L; i < R + 1; i++)
{
if (isPrime(i))
primes.Add(i);
}
for(int i = 0; i < primes.Count; i++)
{
// For every prime number i
// check if an edge
// can be made.
for(int j = i + 1; j < primes.Count; j++)
{
int a = primes[i];
int b = primes[j];
// For every prime number
// i check if an edge can
// be made from i to j.
if (valid(a, b))
{
// If edge is possible then
// insert in the adjacency
// list
lis[a].Add(b);
lis[b].Add(a);
}
}
}
}
// Function to count distance
static void bfs(int src)
{
List q = new List();
q.Add(src);
vis[src] = 1;
dis[src] = 0;
while (q.Count != 0)
{
int curr = q[0];
q.RemoveAt(0);
foreach(int x in lis[curr])
{
// If unvisited push onto queue
// and mark visited as 1 and
// add the distance of curr+1.
if (vis[x] == 0)
{
vis[x] = 1;
q.Add(x);
dis[x] = dis[curr] + 1;
}
}
}
}
// Driver code
static void Main()
{
for(int i = 0; i < 100001; i++)
{
lis.Add(new List());
}
int N = 4;
makePrimes(N);
int A = 1033;
int B = 8179;
// Call bfs traversal
// with root as node A
bfs(A);
if (dis[B] == -1)
{
// Indicates not possible
Console.Write(-1);
}
else
{
Console.Write(dis[B]);
}
}
} 输出：
6时间复杂度： O(10 2N )
辅助空间复杂度： O(10 5 )