Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Program to count number of ways we can place nonoverlapping edges to connect all nodes in C++
Suppose we have a number n that is representing the number of nodes that are placed circularly. We have to find the number of ways we can place n / 2 edges such that every node is connected by an edge, and that edges does not intersect with each other. If the answer is very large then return result mod 10^9 + 7.
So, if the input is like n = 4, then the output will be 2, as we can group them like below −
To solve this, we will follow these steps −
Define an array dp of size (n/2 + 1)
dp[0] := 1, dp[1] := 1
m := 10^9+7
for initialize i := 2, when i <= n / 2, update (increase i by 1), do −
high := i
dp[i] := 0
for initialize j := 1, when j <= high / 2, update (increase j by 1), do −
dp[i] := (dp[i] + (2 * dp[j - 1] * dp[high - j])) mod m
if high % 2 is non-zero, then −
dp[i] := (dp[i] + (dp[(high - 1) / 2] * dp[(high - 1) / 2])) mod m
dp[i] := dp[i] mod m
return dp[n / 2]
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(int n) {
vector<long long> dp(n / 2 + 1);
dp[0] = 1;
dp[1] = 1;
int m = 1000000007;
for (int i = 2; i <= n / 2; i++) {
int high = i;
dp[i] = 0;
for (int j = 1; j <= high / 2; j++) {
dp[i] = (dp[i] + (2 * dp[j - 1] * dp[high - j])) % m;
}
if (high % 2) dp[i] = (dp[i] + (dp[(high - 1) / 2] * dp[(high - 1) / 2])) % m;
dp[i] %= m;
}
return dp[n / 2];
}
main(){
int n = 4;
cout << solve(n);
}Input
4
Output
2
63 Views
