我已经为表达式树的图形着色编写了一个C ++代码（这是我的项目模拟的一部分，与寄存器分配有关）。
我需要以图形方式显示它。我的代码采用中缀表达式并创建表达式树，然后生成它的邻接矩阵并为其着色。
对于任何给定的表达式，我需要以图形方式显示在图像{not getting upload>中。
请建议我使用这种C ++代码的方法。
这是我的代码：
#include <iostream>
#include <string.h>
#include <ctype.h>
#include<conio.h>
#include<stdlib.h>
using namespace std;
struct node{
char data;
int num;
node *left;
node *right;
};
int top1=-1;
node *arr\[35\],*root1\[35\];
int x\[50\],n,tot_colors=0,c=0;
int adj\[20\]\[20\];
class infix
{
private :
char target\[50\], stack\[50\] ;
char *s, *t ;
int top ;
public :
infix( ) ;
void setexpr ( char *str ) ;
void push ( char c ) ;
char pop( ) ;
void convert( ) ;
int priority ( char c ) ;
char* show( ) ;
friend string infixToPostfix(char exp\[\]);
} ;
infix :: infix( )
{
top = -1 ;
strcpy ( target, "" ) ;
strcpy ( stack, "" ) ;
t = target ;
s = "" ;
}
void infix :: setexpr ( char *str )
{
s = str ;
}
void infix :: push ( char c )
{
if ( top == 50 )
cout << "\nStack is full\n" ;
else
{
top++ ;
stack\[top\] = c ;
}
}
char infix :: pop( )
{
if ( top == -1 )
{
cout << "\nStack is empty\n" ;
return -1 ;
}
else
{
char item = stack\[top\] ;
top-- ;
return item ;
}
}
void infix :: convert( )
{
while ( *s )
{
if ( *s == ' ' || *s == '\t' )
{
s++ ;
continue ;
}
if ( isdigit ( *s ) || isalpha ( *s ) )
{
while ( isdigit ( *s ) || isalpha ( *s ) )
{
*t = *s ;
s++ ;
t++ ;
}
}
if ( *s == '(' )
{
push ( *s ) ;
s++ ;
}
char opr ;
if ( *s == '*' || *s == '+' || *s == '/' || *s == '%' || *s == '-' || *s == '$' || *s=='=' )
{
if ( top != -1 )
{
opr = pop( ) ;
while ( priority ( opr ) >= priority ( *s ) )
{
*t = opr ;
t++ ;
opr = pop( ) ;
}
push ( opr ) ;
push ( *s ) ;
}
else
push ( *s ) ;
s++ ;
}
if ( *s == ')' )
{
opr = pop( ) ;
while ( ( opr ) != '(' )
{
*t = opr ;
t++ ;
opr = pop( ) ;
}
s++ ;
}
}
while ( top != -1 )
{
char opr = pop( ) ;
*t = opr ;
t++ ;
}
*t = '\0' ;
}
int infix :: priority ( char c )
{ if(c=='=')
return -1;
if ( c == '$' )
return 3 ;
if ( c == '*' || c == '/' || c == '%' )
return 2 ;
else
{
if ( c == '+' || c == '-' )
return 1 ;
else
return 0 ;
}
}
char* infix :: show( )
{
return target ;
}
string infixToPostfix(char exp\[\])
{
infix q;
q.setexpr ( exp ) ;
q.convert( ) ;
string a=q.show();
return a ;
}
//----------------------------------------------------------------------------------------------------------------------
int r(char inputchar) //for checking symbol is operand or operator for tree
{
if(inputchar=='+' || inputchar=='-' || inputchar=='*' || inputchar=='/' || inputchar=='='|| inputchar=='$')
return(-1);
else if(inputchar>='a' || inputchar<='z')
return(1);
else if(inputchar>='A' || inputchar<='Z')
return(1);
else
return(-99); //for error
}
//it is used for inserting an single element in a tree, i.e. is pushing of single element. FOR TREE
void push1(node *tree){
top1++;
arr\[top1\]=tree;
}
node *pop1(){
top1--;
return(arr\[top1+1\]);
}
//FOR TREE
void create_expr_tree(string suffix)
{
char symbol;
node *newl,*ptr1,*ptr2;
int flag; //flag=-1 when operator and flag=1 when operand
symbol = suffix\[0\]; //Read the first symbol from the postfix expr.
for(int i=1;symbol!=NULL;i++)
{ //continue till end of the expr.
flag=r(symbol); //check symbol is operand or operator.
if(flag == 1) //if symbol is operand.
{
newl = new node;
//new1->num=0;
newl->data = symbol;
newl->left = NULL;
newl->right = NULL;
push1(newl);
}
else{ //If the symbol is operator//pop two top elements.
ptr1=pop1();
ptr2=pop1();
newl = new node;
newl->data = symbol;
newl->left = ptr2;
newl->right = ptr1;
push1(newl);
}
symbol=suffix\[i\];
}
}
void inOrder(node *tree){
if( tree!=NULL){
inOrder( tree->left);
cout<< tree->data;
inOrder(tree->right);
}
}
node* postfixToTree(string postfix)
{
create_expr_tree(postfix);
return arr\[0\];
}
//----------------------------------------------------------------------------------------------------/+
void bfs(node *root)
{
if(c<n)
{
root->num=c++;
cout<<root->data<<"\t"<<root->num<<"\n";
if(root->left!=NULL)
{
bfs(root->left);
}
if(root->right!=NULL)
{
bfs(root->right);
}
}
}
void adjgen(node *root)
{
node *n1,*n2;
n1=root->left;
n2=root->right;
if(root->left!=NULL)
{ adj\[root->num\]\[n1->num\]=1;
adj\[n1->num\]\[root->num\]=1;
adjgen(root->left);
}
if(root->right!=NULL)
{ adj\[root->num\]\[n2->num\]=1;
adj\[n2->num\]\[root->num\]=1;
adjgen(root->right);
}
}
//-------------------------------------------------gc
int graph_color(int k)
{
int i;
x\[k\]=1; //coloring vertex with color1
for(i=0;i<k;i++){ //checking all k-1 vertices-backtracking
if(adj\[i\]\[k\]!=0 && x\[k\]==x\[i\]) //if connected and has same color
{ x\[k\]=x\[i\]+1;//assign higher color than x\[i\]
if(x\[k\]>tot_colors)
{
cout<<"\nThe given graph is not "<<tot_colors<<"-colorable!!!";
exit(0);
}
}
}
}
int o=0;
void show_color(node *root,int a\[\])
{ cout<<root->data<<"\t"<<a\[o++\]<<"\n";
if(root->left!=NULL)
{
show_color(root->left,a);
}
if(root->right!=NULL)
{
show_color(root->right,a);
}
}
int main( )
{
char expr\[50\];
int i,j;
//string p1;
cout << "\nEnter an expression in infix form: " ;
cin>>expr ;
string p1=infixToPostfix(expr);
n=p1.size();
cout << "\nThe postfix expression is: " <<p1<<endl;
root1\[0\]=postfixToTree(p1);
cout<<"\nIn-Order Traversal : ";
inOrder(root1\[0\]);
cout<<"\nBFS\n";
bfs(root1\[0\]);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
adj\[i\]\[j\]=0;
}
}
adjgen(root1\[0\]);
cout<<"adjagen\n";
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
cout<<adj\[i\]\[j\];
}
cout<<"\n";
}
//-----------------------------------------------------GC
cout<<"\nEnter the no. of colors available(no. of registers available): ";
cin>>tot_colors;
int fl=1;
for(i=0;i<n;i++)
graph_color(i);
cout<<"\nNODE\tCOLOR\n";
show_color(root1\[0\],x);
// for(i=0;i<n;i++) //displaying color of each vertex
//cout<<"\nVertex\["<<i<<"\] : "<<x\[i\];
}
图表必须如下：
a
/ \
= +
/ \
b c
应该根据上述程序用图形表示。节点应该是圆的
答案 0 :(得分：0)
如果您需要GUI，请留意某些库，例如SDL和SDL_gfx将其排序。或者，尝试使用windows.h附带的Windows API。