有向圖的強連通分支演算法kosaraju（C語言實現)
阿新 • • 發佈：2019-02-10
/* Kosaraju求強連通分量鄰接矩陣 */
#include "stdio.h"
#include "stdlib.h"
#define Vexnum 100
int map1[Vexnum][Vexnum]={0};
int nmap[Vexnum][Vexnum]={0};
int visited[Vexnum];
/*這部分是棧的定義和操作，可直接使用*/
struct StackSq{
int *stack1;
int top;
int MaxSize;
};
struct StackSq s;
void InitStack(struct StackSq *s)
{
s->MaxSize=100;/*置棧空間初始最大長度為100*/
s->stack1=(int *)malloc(10*sizeof(int));/*動態儲存空間分配*/
s->top=-1;/*置棧為空*/
}
//1
void Push(struct StackSq* S,int x)
{
//if(S->top==S->MaxSize-1)againMalloc(S);/*若棧空間用完則重新分配更大的儲存空間*/
S->top++;/*棧頂指標後移一個位置*/
S->stack1[S->top]=x;/*將新元素插入到棧頂*/
}
//2
int Pop(struct StackSq* S)
{
if(S->top==-1){
printf("棧空，無元素出棧！\n");
exit(1);
}
S->top--;
return S->stack1[S->top+1];
}
//3
int Peek(struct StackSq* s)
{
if(s->top==-1){
printf("棧空，無元素出棧！\n");
exit(1);
}
return s->stack1[s->top];
}
int EmptyStack(struct StackSq* S)
{
if(S->top==-1) return 1;
else return 0;
}
int N; /*頂點數目*/
int DFS1(int v) /* 深度遍歷有向圖G*/
{
visited[v] = 1;
for (int i = 1;i <= N;i++)
if (!visited[i] && map1[v][i])
DFS1( i );
Push(&s,v); /*這步很妙,時間順序*/
return 0;
}
int DFS2( int v ) /* 深度遍歷有向圖G的逆圖*/
{
visited[v] = 1;
printf( "%d,", v );
for (int i = 1;i <= N;i++)
if ( !visited[i] && nmap[v][i] )
DFS2( i );
return 0;
}
int kosaraju()
{
int i;
while (!EmptyStack(&s))
Pop(&s);
for(i=1;i<=N;i++)
visited[i]=0;
for (int i = 1;i <= N;i++)
if (!visited[i]) DFS1( i );
int t = 0;
for(i=1;i<=N;i++)
visited[i]=0;
while (!EmptyStack(&s))
{
int v = Peek(&s);
Pop(&s);
printf("{");
if (!visited[v])
{
t++;
DFS2( v );
}
printf("}");
}
return t;
}
#include "stdio.h"
#include "stdlib.h"
#define Vexnum 100
int map1[Vexnum][Vexnum]={0};
int nmap[Vexnum][Vexnum]={0};
int visited[Vexnum];
/*這部分是棧的定義和操作，可直接使用*/
struct StackSq{
int *stack1;
int top;
int MaxSize;
};
struct StackSq s;
void InitStack(struct StackSq *s)
{
s->MaxSize=100;/*置棧空間初始最大長度為100*/
s->stack1=(int *)malloc(10*sizeof(int));/*動態儲存空間分配*/
s->top=-1;/*置棧為空*/
}
//1
void Push(struct StackSq* S,int x)
{
//if(S->top==S->MaxSize-1)againMalloc(S);/*若棧空間用完則重新分配更大的儲存空間*/
S->top++;/*棧頂指標後移一個位置*/
S->stack1[S->top]=x;/*將新元素插入到棧頂*/
}
//2
int Pop(struct StackSq* S)
{
if(S->top==-1){
printf("棧空，無元素出棧！\n");
exit(1);
}
S->top--;
return S->stack1[S->top+1];
}
//3
int Peek(struct StackSq* s)
{
if(s->top==-1){
printf("棧空，無元素出棧！\n");
exit(1);
}
return s->stack1[s->top];
}
int EmptyStack(struct StackSq* S)
{
if(S->top==-1) return 1;
else return 0;
}
int N; /*頂點數目*/
int DFS1(int v) /* 深度遍歷有向圖G*/
{
visited[v] = 1;
for (int i = 1;i <= N;i++)
if (!visited[i] && map1[v][i])
DFS1( i );
Push(&s,v); /*這步很妙,時間順序*/
return 0;
}
int DFS2( int v ) /* 深度遍歷有向圖G的逆圖*/
{
visited[v] = 1;
printf( "%d,", v );
for (int i = 1;i <= N;i++)
if ( !visited[i] && nmap[v][i] )
DFS2( i );
return 0;
}
int kosaraju()
{
int i;
while (!EmptyStack(&s))
Pop(&s);
for(i=1;i<=N;i++)
visited[i]=0;
for (int i = 1;i <= N;i++)
if (!visited[i]) DFS1( i );
int t = 0;
for(i=1;i<=N;i++)
visited[i]=0;
while (!EmptyStack(&s))
{
int v = Peek(&s);
Pop(&s);
printf("{");
if (!visited[v])
{
t++;
DFS2( v );
}
printf("}");
}
return t;
}