M coloring problem

Given an undirected graph and a number m, determine if the graph can be coloured with at most m colours such that no two adjacent vertices of the graph are colored with the same color.
Input : in input we have taken a undirected graph in which user has to enter no of vertices and edges between vertices also we have to provide a number m which is the maximum number of colors that can be used.
Output:In output the array is displayed of all the pattern in which we can color the graph .If there is no possiblity then program is simply terminated
/* Problem is solved by using backtracking . in backtracking we deal problem with searching for a set of solutions or which ask for an optimal solution that satisfiessome constraits can be solves using backtracking formulation 
//Program in C++
#include<stdio.h>

int nextvalue(int k,int x[],int n,int a[][20],int m);

int mcoloring(int k,int x[],int n,int a[][20],int m)
{
	int i;
	do
	{
		nextvalue(k,x,n,a,m);
		if(x[k]==0)
		     return 0;
		if(k==n)
		{
			for(i=1;i<=n;i++)
			printf("%d\t",x[i]);
			printf("\n");
		}
		else
		        mcoloring(k+1,x,n,a,m);
	}while(1);
}
// function for genertaing next color
int nextvalue(int k,int x[],int n,int a[][20],int m)
{
	int j;
	do
	{
		x[k]=(x[k]+1)%(m+1);// next higher color
		
		if(x[k]==0)
		        return 0;// when all colors are used
		for(j=1;j<=n;j++)
		{
			if((a[k][j]!=0)&&(x[k]==x[j]))// checks if the color is distinct from adjacent colors
			      break;
		}
		if(j==n+1)// that means new color found
		        return 0;
	}while(1);
}

int main()
{
	int n,ne,k,x[20],ar[20][20],i,j,a,b,m;
	
	printf("enter no of vertices: ");
	scanf("%d",&n);
	
	printf("enter total no of edges: ");
	scanf("%d",&ne);
	
	printf("m= ");
	scanf("%d",&m);
	
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
	               ar[i][j]=0;
	               
	        x[i]=0;
	}
		      
		             
	for(i=1;i<=ne;i++)
	{
		printf("enter first terminal: ");
		scanf("%d",&a);
		
		printf("enter second terminal: ");
		scanf("%d",&b);
		
		ar[a][b]=1;
		ar[b][a]=1;
	}
	
	mcoloring(1,x,n,ar,m);	
}
/*
Time Complexity: O(m^V). 
There are total O(m^V) combination of colors. So time complexity is O(m^V).
Space Complexity: O(V). 
It will require O(V) space.
Test case 1
enter no of vertices: 5
enter total no of edges: 5
m= 3
enter first terminal: 1
enter second terminal: 2
enter first terminal: 2
enter second terminal: 3
enter first terminal: 3
enter second terminal: 4
enter first terminal: 4
enter second terminal: 5
enter first terminal: 5
enter second terminal: 1
1       2       1       2       3
1       2       1       3       2
1       2       3       1       2
1       2       3       1       3
1       2       3       2       3
1       3       1       2       3
1       3       1       3       2
1       3       2       1       2
1       3       2       1       3
1       3       2       3       2
2       1       2       1       3
2       1       2       3       1
2       1       3       1       3
2       1       3       2       1
2       1       3       2       3
2       3       1       2       1
2       3       1       2       3
2       3       1       3       1
2       3       2       1       3
2       3       2       3       1
3       1       2       1       2
3       1       2       3       1
3       1       2       3       2
3       1       3       1       2
3       1       3       2       1
3       2       1       2       1
3       2       1       3       1
3       2       1       3       2
3       2       3       1       2
3       2       3       2       1
Test case 2
enter no of vertices: 3
enter total no of edges: 1
m= 2
enter first terminal: 1
enter second terminal: 2
1       2       1
1       2       2
2       1       1
2       1       2

*/