给定一个由N 个字符组成的数组arr[] ，任务是生成N 个节点和(N – 1) 条边的图，使得每个节点i与字符arr[i]相关联，并且没有两个相邻节点具有相同的值.如果可以制作这样的图，则用边对打印“可能” 。否则，打印“不可能” 。
例子：
Input: N = 5, arr[] = {‘a’, ‘b’, ‘a’, ‘b’, ‘c’}
Output: “Possible”
1 – 2
1 – 4
1 – 5
5 – 3
Explanation:
One possible graph that can be constructed to satisfy the given conditions is as follows:
Input: N = 3, arr[] = {‘z’, ‘z’, ‘z’}
Output: “Not Possible”
方法：要构造一个图，使得相邻节点都没有相同的值，其想法是检查是否至少存在两个唯一值。如果发现为真，则可以构建这样的图。请按照以下步骤操作：
- 检查每个节点上存在的所有值，如果所有节点值都相同，则无法构建图形。
- 如果任何两个值不同，总会有一种方法来构造这样的图。
- 现在，选择任意两个唯一值，将所有其他值的出现连接到除值本身之外的第一个唯一值。
- 存储第一个唯一值的出现索引，除了第一次出现，并将所有这些索引连接到第二个唯一值。
- 这样，总有一种方法可以构造这样一个具有(N – 1) 条边的图。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function that prints the edges of
// the generated graph
void printConnections(
vector > store,
vector ind, int ind1)
{
// First print connections
// stored in store[]
for (auto pr : store) {
cout << pr.first << " "
<< pr.second << "\n";
}
// Check if there is more than one
// occurrence of 1st unique element
if (ind.size() != 0) {
// Print all other occurrence
// of 1st unique element with
// second unique element
for (auto x : ind) {
cout << ind1 << " "
<< x + 1 << "\n";
}
}
}
// Function to construct the graph such
// that the every adjacent nodes have
// different value
void constructGraph(char arr[], int N)
{
vector ind;
// Stores pair of edges formed
vector > store;
// Stores first unique occurrence
char x = arr[0];
int count = 0, ind1;
for (int i = 1; i <= N - 1; ++i) {
// Check for the second
// unique occurrence
if (arr[i] != x) {
// Store indices of 2nd
// unique occurrence
ind1 = i + 1;
// To check if arr has only
// 1 unique element or not
count++;
// Store the connections of all
// unique elements with Node 1
store.push_back({ 1, i + 1 });
}
// If value at node (i + 1) is
// same as value at Node 1 then
// store its indices
else {
ind.push_back(i);
}
}
// If count is zero then it's not
// possible to construct the graph
if (count == 0) {
cout << "Not Possible";
}
// If more than 1 unique
// element is present
else {
cout << "Possible"
<< "\n";
// Print the edges
printConnections(store, ind, ind1);
}
}
// Driver Code
int main()
{
int N = 5;
// Given array having node values
char arr[] = { 'a', 'b', 'a', 'b', 'c' };
// Function Call
constructGraph(arr, N);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
// Function that prints the edges of
// the generated graph
static void printConnections(Vector store,
Vector ind,
int ind1)
{
// First print connections
// stored in store[]
for (pair pr : store)
{
System.out.print(pr.first + " " +
pr.second + "\n");
}
// Check if there is more than one
// occurrence of 1st unique element
if (ind.size() != 0)
{
// Print all other occurrence
// of 1st unique element with
// second unique element
for (int x : ind)
{
System.out.print(ind1 + " " +
(x + 1) + "\n");
}
}
}
// Function to conthe graph such
// that the every adjacent nodes have
// different value
static void constructGraph(char arr[],
int N)
{
Vector ind = new Vector<>();
// Stores pair of edges formed
Vector store = new Vector<>();
// Stores first unique occurrence
char x = arr[0];
int count = 0;
int ind1=-1;
for (int i = 1; i <= N - 1; ++i)
{
// Check for the second
// unique occurrence
if (arr[i] != x)
{
// Store indices of 2nd
// unique occurrence
ind1 = i + 1;
// To check if arr has only
// 1 unique element or not
count++;
// Store the connections of all
// unique elements with Node 1
store.add(new pair(1, i + 1 ));
}
// If value at node (i + 1) is
// same as value at Node 1 then
// store its indices
else
{
ind.add(i);
}
}
// If count is zero then it's not
// possible to conthe graph
if (count == 0)
{
System.out.print("Not Possible");
}
// If more than 1 unique
// element is present
else
{
System.out.print("Possible" +
"\n");
// Print the edges
printConnections(store,
ind, ind1);
}
}
// Driver Code
public static void main(String[] args)
{
int N = 5;
// Given array having
// node values
char arr[] = {'a', 'b',
'a', 'b', 'c'};
// Function Call
constructGraph(arr, N);
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program for the above approach
# Function that prints the edges of
# the generated graph
def printConnections(store, ind, ind1):
# First print connections
# stored in store[]
for pr in store:
print(pr[0], pr[1])
# Check if there is more than one
# occurrence of 1st unique element
if (len(ind) != 0):
# Print all other occurrence
# of 1st unique element with
# second unique element
for x in ind:
print(ind1, x + 1)
# Function to construct the graph such
# that the every adjacent nodes have
# different value
def constructGraph(arr, N):
ind = []
# Stores pair of edges formed
store = []
# Stores first unique occurrence
x = arr[0]
count, ind1 = 0, 0
for i in range(1, N):
# Check for the second
# unique occurrence
if (arr[i] != x):
# Store indices of 2nd
# unique occurrence
ind1 = i + 1
# To check if arr has only
# 1 unique element or not
count += 1
# Store the connections of all
# unique elements with Node 1
store.append([1, i + 1])
# If value at node (i + 1) is
# same as value at Node 1 then
# store its indices
else:
ind.append(i)
# If count is zero then it's not
# possible to construct the graph
if count == 0:
print("Not Possible")
else:
# If more than 1 unique
# element is present
print("Possible")
# Print the edges
printConnections(store, ind, ind1)
# Driver Code
if __name__ == '__main__':
N = 5
# Given array having node values
arr = [ 'a', 'b', 'a', 'b', 'c' ]
# Function Call
constructGraph(arr, N)
# This code is contributed by mohit kumar 29C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
public class pair
{
public int first,
second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
// Function that prints the edges of
// the generated graph
static void printConnections(List store,
List ind,
int ind1)
{
// First print connections
// stored in store[]
foreach (pair pr in store)
{
Console.Write(pr.first + " " +
pr.second + "\n");
}
// Check if there is more than one
// occurrence of 1st unique element
if (ind.Count != 0)
{
// Print all other occurrence
// of 1st unique element with
// second unique element
foreach (int x in ind)
{
Console.Write(ind1 + " " +
(x + 1) + "\n");
}
}
}
// Function to conthe graph such
// that the every adjacent nodes have
// different value
static void constructGraph(char []arr,
int N)
{
List ind = new List();
// Stores pair of edges formed
List store = new List();
// Stores first unique occurrence
char x = arr[0];
int count = 0;
int ind1=-1;
for (int i = 1; i <= N - 1; ++i)
{
// Check for the second
// unique occurrence
if (arr[i] != x)
{
// Store indices of 2nd
// unique occurrence
ind1 = i + 1;
// To check if arr has only
// 1 unique element or not
count++;
// Store the connections of all
// unique elements with Node 1
store.Add(new pair(1, i + 1 ));
}
// If value at node (i + 1) is
// same as value at Node 1 then
// store its indices
else
{
ind.Add(i);
}
}
// If count is zero then it's not
// possible to conthe graph
if (count == 0)
{
Console.Write("Not Possible");
}
// If more than 1 unique
// element is present
else
{
Console.Write("Possible" +
"\n");
// Print the edges
printConnections(store,
ind, ind1);
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 5;
// Given array having
// node values
char []arr = {'a', 'b',
'a', 'b', 'c'};
// Function Call
constructGraph(arr, N);
}
}
// This code is contributed by gauravrajput1 输出：
Possible
1 2
1 4
1 5
5 3
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