Description of the problem:
Given an undirected graph, m different colors. So that the coloring of each of two vertices of G different colors each edge, if a minimum of m FIG colors in order to make FIG. 2 different colors of each vertices connected by edges, so that the number m for the number of color charts. Problems FIG m seeking a number of colors m may be referred to FIG optimization coloring.
2 algorithm design
By adjacency matrix A represents a non-connected graph G to FIG = (V, E).
If the connected edges is present, a [i] [j] = 1, or a [i] [j] = 0.
Integers 1,2,3 .. . m is used to denote a full height of n + m is a tree.
Each node in the i-th layer solution space tree has m sons, each corresponding to a son of one of the possible x [i] in the m coloration.
The n + 1 layer is a leaf node.
In the algorithm Backtrack,
When i> n, a search algorithm to the leaf node, obtain a new coloration schemes m, the current can find the program tree colored by 1 m.
When i <= n, the current node Z is extended internal node in the solution space. The node has x [i] = 1,2,3 .. . m co m son nodes. For each of a son node of the current node Z of the extension, which checks the feasibility of the OK function, and the depth-first manner recursively subtree search feasible, infeasible, or cut subtree.
Algorithm Description:
#include <iostream>
using namespace std;
class Color{
friend int mColoring(int,int,int* *);
private:
bool Ok(int k);
void Backtrack(int t);
int n,
m,
* *a,
* x;
long sum;
};
bool Color::Ok(int k)
{
for(int j=1;j<=n;j++)
if((a[k][j] == 1)&&(x[j] == x[k]))
return false;
return true;
}
void Color::Backtrack(int t)
{
if(t>n)
{
sum++;
for(int i=1;i<=n;i++)
cout<<x[i]<<" ";
cout<<endl;
}
else
{
for(int i=1;i<=m;i++)
x[t] = i;
if(Ok(t))
Backtrack(t+1);
x[t] = 0;
}
}
int mColoring(int n,int m,int * * a)
{
Color X;
X.n = n;
X.m = m;
X.a = a;
X.sum = 0;
int * p = new int [n+1];
for(int i=0;i<=n;i++)
p[i] = 0;
X.x = p;
X.Backtrack(2);
delete [] p;
return X.sum;
}
int main()
{
int n,m;
cout<<"请输入想要确定的m着色图中m的值："<<endl;
cin>>m;
cout<<endl<<"共有n个结点，n的值为："<<endl;
cin>>n;
int **arr = new int [n][n];
for(int i=0;i<n;i++)
{
cout<<"请输入连接矩阵的第一行的数（0-1）"<<endl;
for(int j=0;j<n;j++)
{
cin>>arr[i][j];
}
}
mColoring(n,m,arr);
return 0;
}operation result:
--------------------Configuration: test102501 - Win32 Debug--------------------
Compiling...
test.cpp
E:\study file\ACM\test102501\test.cpp(63) : error C2540: non-constant expression as array bound
E:\study file\ACM\test102501\test.cpp(63) : error C2440: 'initializing' : cannot convert from 'int (*)[1]' to 'int ** '
Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
Error executing cl.exe.
test.obj - 2 error(s), 0 warning(s)Or an array of pointers will not pass parameters ... time to see, mark it
Reproduced in: https: //my.oschina.net/u/204616/blog/545461