给定一个有N个顶点和N 个边的无向图。没有两条边连接同一对顶点。三角形是一组三个不同的顶点，每对这些顶点都由一条边连接，即如果图包含边(u, v) , (v, w) ，则三个不同的顶点u , v和w是三角形和(v, u) 。
任务是找到需要添加到给定图中的最少边数，以使该图至少包含一个三角形。
例子：
Input:
1
/ \
2 3
Output: 1
Input:
1 3
/ /
2 4
Output: 2
方法：初始化ans = 3 ，这是形成三角形所需的最大边数。现在，对于每个可能的顶点三元组，有四种情况：
- 情况 1：如果存在节点i 、 j和k使得有来自(i, j) 、 (j, k)和(k, i) 的边，则答案为0 。
- 情况 2：如果存在节点i 、 j和k使得只有两对顶点连接，则需要一条边来形成三角形。所以更新ans = min(ans, 1) 。
- 情况 3：否则，如果只有一对顶点连接，则ans = min(ans, 2) 。
- 情况 4：当没有边时ans = min(ans, 3) 。
最后打印答案。
下面是上述方法的实现。
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the minimum number
// of edges that need to be added to
// the given graph such that it
// contains at least one triangle
int minEdges(vector > v, int n)
{
// adj is the adjacency matrix such that
// adj[i][j] = 1 when there is an
// edge between i and j
vector > adj;
adj.resize(n + 1);
for (int i = 0; i < adj.size(); i++)
adj[i].resize(n + 1, 0);
// As the graph is undirected
// so there will be an edge
// between (i, j) and (j, i)
for (int i = 0; i < v.size(); i++) {
adj[v[i].first][v[i].second] = 1;
adj[v[i].second][v[i].first] = 1;
}
// To store the required
// count of edges
int edgesNeeded = 3;
// For every possible vertex triplet
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
for (int k = j + 1; k <= n; k++) {
// If the vertices form a triangle
if (adj[i][j] && adj[j][k] && adj[k][i])
return 0;
// If no edges are present
if (!(adj[i][j] || adj[j][k] || adj[k][i]))
edgesNeeded = min(edgesNeeded, 3);
else {
// If only 1 edge is required
if ((adj[i][j] && adj[j][k])
|| (adj[j][k] && adj[k][i])
|| (adj[k][i] && adj[i][j])) {
edgesNeeded = 1;
}
// Two edges are required
else
edgesNeeded = min(edgesNeeded, 2);
}
}
}
}
return edgesNeeded;
}
// Driver code
int main()
{
// Number of nodes
int n = 3;
// Storing the edges in a vector of pairs
vector > v = { { 1, 2 }, { 1, 3 } };
cout << minEdges(v, n);
return 0;
} Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class Pair
{
V first;
E second;
Pair(V first, E second)
{
this.first = first;
this.second = second;
}
}
class GFG
{
// Function to return the minimum number
// of edges that need to be added to
// the given graph such that it
// contains at least one triangle
static int minEdges(Vector> v, int n)
{
// adj is the adjacency matrix such that
// adj[i][j] = 1 when there is an
// edge between i and j
int[][] adj = new int[n + 1][n + 1];
// As the graph is undirected
// so there will be an edge
// between (i, j) and (j, i)
for (int i = 0; i < v.size(); i++)
{
adj[v.elementAt(i).first]
[v.elementAt(i).second] = 1;
adj[v.elementAt(i).second]
[v.elementAt(i).first] = 1;
}
// To store the required
// count of edges
int edgesNeeded = 0;
// For every possible vertex triplet
for (int i = 1; i <= n; i++)
{
for (int j = i + 1; j <= n; j++)
{
for (int k = j + 1; k <= n; k++)
{
// If the vertices form a triangle
if (adj[i][j] == 1 &&
adj[j][k] == 1 && adj[k][i] == 1)
return 0;
// If no edges are present
if (!(adj[i][j] == 1 ||
adj[j][k] == 1 || adj[k][i] == 1))
edgesNeeded = Math.min(edgesNeeded, 3);
else
{
// If only 1 edge is required
if ((adj[i][j] == 1 && adj[j][k] == 1) ||
(adj[j][k] == 1 && adj[k][i] == 1) ||
(adj[k][i] == 1 && adj[i][j] == 1))
{
edgesNeeded = 1;
}
// Two edges are required
else
edgesNeeded = Math.min(edgesNeeded, 2);
}
}
}
}
return edgesNeeded;
}
// Driver Code
public static void main(String[] args)
{
// Number of nodes
int n = 3;
// Storing the edges in a vector of pairs
Vector> v = new Vector<>(Arrays.asList(new Pair<>(1, 2),
new Pair<>(1, 3)));
System.out.println(minEdges(v, n));
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 implementation of the approach
# Function to return the minimum number
# of edges that need to be added to
# the given graph such that it
# contains at least one triangle
def minEdges(v, n) :
# adj is the adjacency matrix such that
# adj[i][j] = 1 when there is an
# edge between i and j
adj = dict.fromkeys(range(n + 1));
# adj.resize(n + 1);
for i in range(n + 1) :
adj[i] = [0] * (n + 1);
# As the graph is undirected
# so there will be an edge
# between (i, j) and (j, i)
for i in range(len(v)) :
adj[v[i][0]][v[i][1]] = 1;
adj[v[i][1]][v[i][0]] = 1;
# To store the required
# count of edges
edgesNeeded = 3;
# For every possible vertex triplet
for i in range(1, n + 1) :
for j in range(i + 1, n + 1) :
for k in range(j + 1, n + 1) :
# If the vertices form a triangle
if (adj[i][j] and adj[j][k] and adj[k][i]) :
return 0;
# If no edges are present
if (not (adj[i][j] or adj[j][k] or adj[k][i])) :
edgesNeeded = min(edgesNeeded, 3);
else :
# If only 1 edge is required
if ((adj[i][j] and adj[j][k])
or (adj[j][k] and adj[k][i])
or (adj[k][i] and adj[i][j])) :
edgesNeeded = 1;
# Two edges are required
else :
edgesNeeded = min(edgesNeeded, 2);
return edgesNeeded;
# Driver code
if __name__ == "__main__" :
# Number of nodes
n = 3;
# Storing the edges in a vector of pairs
v = [ [ 1, 2 ], [ 1, 3 ] ];
print(minEdges(v, n));
# This code is contributed by kanugargngC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class Pair
{
public int first;
public int second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
class GFG
{
// Function to return the minimum number
// of edges that need to be added to
// the given graph such that it
// contains at least one triangle
static int minEdges(List v, int n)
{
// adj is the adjacency matrix such that
// adj[i,j] = 1 when there is an
// edge between i and j
int[,] adj = new int[n + 1, n + 1];
// As the graph is undirected
// so there will be an edge
// between (i, j) and (j, i)
for (int i = 0; i < v.Count; i++)
{
adj[v[i].first,v[i].second] = 1;
adj[v[i].second,v[i].first] = 1;
}
// To store the required
// count of edges
int edgesNeeded = 0;
// For every possible vertex triplet
for (int i = 1; i <= n; i++)
{
for (int j = i + 1; j <= n; j++)
{
for (int k = j + 1; k <= n; k++)
{
// If the vertices form a triangle
if (adj[i, j] == 1 &&
adj[j, k] == 1 && adj[k, i] == 1)
return 0;
// If no edges are present
if (!(adj[i, j] == 1 ||
adj[j, k] == 1 || adj[k, i] == 1))
edgesNeeded = Math.Min(edgesNeeded, 3);
else
{
// If only 1 edge is required
if ((adj[i, j] == 1 && adj[j, k] == 1) ||
(adj[j, k] == 1 && adj[k, i] == 1) ||
(adj[k, i] == 1 && adj[i, j] == 1))
{
edgesNeeded = 1;
}
// Two edges are required
else
edgesNeeded = Math.Min(edgesNeeded, 2);
}
}
}
}
return edgesNeeded;
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int n = 3;
// Storing the edges in a vector of pairs
List v = new List();
v.Add(new Pair(1, 2));
v.Add(new Pair(1, 3));
Console.WriteLine(minEdges(v, n));
}
}
// This code is contributed by PrinciRaj1992 输出：
1