void EDIT( )
{
STACK S; char c ;
MAKENULL ( S );
cin.get(c) ;
while ( c != '\n' )
{
if ( c == '#' )
POP ( S );
else if ( c == '@' )
MAKENULL ( S );
else
PUSH ( c , S );
cin.get(c) ;
}
while(!EMPTY(S))
{
cout<<TOP(S);
POP(S);
}
cout<<endl;
}
boolean Correct(char ext[], int n)
{
STACK S;
char x;
int j=0;
MAKENULL(S);
while(ext[j])
{
if(ext[j]=='(')
PUSH('(', S);
else if(ext[j]==')')
{
x=TOP(S);
if(x=='(')
POP(S);
else
return FALSE;
}
j++;
}
if(!EMPTY(S))
return FALSE;
else
return TRUE;
}
main() {
LIST L, p, p2;
char ans, element;
MAKENULL(&L);
do {
clrscr();
printf("\n MENU\n");
printf("[1] Insert\n");
printf("[2] Delete\n");
printf("[3] Display\n");
printf("[4] Search\n");
printf("[5] Quit\n\n");
printf("Enter chosen number: ");
ans = toupper(getche());
switch (ans) {
case '1':
/* case 1 has an approximated running time of 3n + 12 */
printf("\n\nEnter an element to insert: ");
element = getche();
p = INS_POS(element, L);
INSERT(element, p);
break;
case '2' :
/* case 2 has an approximated running time of 8n + 10 */
printf("\n\nEnter element to delete: ");
element = getche();
p = LOCATE(element,L);
DELETE(element, p);
break;
case '3' :
/* case 3 has an approximated running time of 3n + 4 */
printf("\n\n");
PRINTLIST(L);
break;
case '4' :
/* case 4 has an approximated running time of 4n + 14 */
printf("\n\nEnter element to search: ");
element = getch();
p = LOCATE(element,L);
SEARCH(element, p);
break;
case '5' :
break;
}
} while (ans != '5');
DELETEALL(p); /* DELETEALL has an approximated runnning time of 4n + 1 */
}
void Topologicalsort( AdjGraph G, int aov[NumVertices] )
{
int v, w, nodes;
EdgeNode *tmp;
EdgeData indegree[NumVertices+1]={0};
QUEUE Q ;
MAKENULL( Q ) ;
// 计算每个顶点的入度
for( v=1; v<=G.n ; ++v )
{
tmp=G.vexlist[v].firstedge;
while(tmp)
{
indegree[tmp->adjvex]++;
tmp=tmp->next;
}
}
// 将入度为0的顶点加入队列
for(v=1; v<=G.n; ++v)
if ( indegree[v] ==0 )
ENQUEUE( v, Q ) ;
nodes = 0 ;
while ( !EMPTY( Q ) )
{
v = FRONT(Q)->element ;
DEQUEUE( Q ) ;
//cout << v <<' ';
aov[nodes]=v;
nodes ++ ; // 已考虑的节点个数加1
// 如果(v, w)是一条边，将w的入度减1，如果w的入度为0，则将w入队
for( w=1; w<=G.n; w++)
{
if(connect(G, v, w))
{
--indegree[w];
if( !(indegree[w]))
ENQUEUE(w,Q) ;
}
}
}
cout<<endl;
if ( nodes < G.n )
cout<<"图中有环路"<<endl;
}
