m Coloring Problem using c or c++

Given an undirected graph and a number m, determine if the graph can be colored with at most m colors such that no two adjacent vertices of the graph are colored with the same color. Here coloring of a graph means the assignment of colors to all vertices.

Input:
1) A 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph. A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0.
2) An integer m which is a maximum number of colors that can be used.

Output:
An array color[V] that should have numbers from 1 to m. color[i] should represent the color assigned to the ith vertex. The code should also return false if the graph cannot be colored with m colors.

Program:

#include <stdio.h>
#include <conio.h>
#include <math.h>

int m,n,x[10]={0},g[10][10]={0};

void main()
{
                int i,j; void mcolor();

                printf("No. of colors available: ");
                scanf("%d",&m);
                printf("No. of regions: ") ;
                scanf("%d",&n) ;
                for(i=1;i<=n;i++)
                                for(j=1;j<=n;j++)
                                {
                                                printf("g[%d][%d] = ",i,j) ;
                                                scanf("%d",&g[i][j]) ;
                                }
                mcolor();

                getch();
}

void mcolor()
{
                int i,k=1,backtrack(int),solution(int) ;
                void printsol() ;

                while(k!=0)
                {
                                x[k]=(x[k]+1)%(m+1) ;
                                printf("x[%d] = %d\n",k,x[k]) ; getch() ;
                                if(x[k]!=0)
                                {
                                                if(backtrack(k))
                                                                continue ;
                                                if(solution(k))
                                                {
                                                                printsol() ;
                                                                return ;
                                                }
                                                else
                                                                k++ ;
                                }
                                else
                                {
                                                printf("%d colors are not sufficient\n",m) ;
                                                return ;
                                }
                }
}

int backtrack(int k)
{
                int j;
                for(j=1;j<k;j++)
                                if(g[j][k]==1 && x[j]==x[k])
                                {
                                                return 1;
                                }
                return 0;
}

int solution(int k)
{
                if(k==n)
                                return 1 ;
                else
                                return 0 ;
}

void printsol()
{
                printf("Solution is : ") ;
                for(int j=1;j<=n;j++)
                                printf("%d ",x[j]);
                printf("\n");
}