2013-02-02 36 views
0

在函數Dijkstra中,我初始化一個優先級隊列nperm,它具有一個信息字段;它告訴節點是否是非永久性的集合的成員。我有一個的「i」的字段,它實際存儲指針(數組索引),以該圖形節點和下一pointer.Initially每個節點是notpermanent set.This的構件是如何初始化完成:試圖實現Dijkstra,但有分段錯誤沒有意義

for (i = graph; hnode[i].next >= 0;) 
{ 
    insertnperm(&nperm, 1, i); 
    distance[i] = 500000; 
    i = hnode[i].next; 
} 

距離[i]基本上是節點i距目前爲止已知的根節點的距離,因此隨着算法的進行,距離被更新,因此應該是基於距離的優先級隊列nperm的順序。 現在按函數順序傳遞一個指向優先級隊列指針的指針,並將該指針的值存儲在另一個相同類型的變量notperm p; p = *pq;中,因爲優先級隊列中的每個節點都有一個字段'i',其中包含圖形節點,我遍歷列表直到最近更新距離的節點,並刪除並將該節點放置在上升優先級隊列中的適當位置。

我使用陳述while ((p->i) != s)遍歷到節點,但出現了分段錯誤,其中我不知道。當我使用p->i訪問「我」時,似乎發生分段錯誤。

任何人都可以向我解釋爲什麼會發生這種情況?

這是全碼:

#include<stdio.h> 
#include<conio.h> 
#define MAXNODES 50 

struct header 
{ 
    int info;  
    int epoint;  
    int tpoint;  
    int next;  
}; 

struct header hnode[MAXNODES];  
int i = 0; 

struct arcs 
{  
    int info;  
    int epoint;  
    int tpoint;  
    int enext;  
    int tnext;  
};  
int j = 0;  
struct arcs anode[MAXNODES]; 

struct node 
{  
    int info;  
    int i;  
    struct node *next;  
};  
typedef struct node *notperm;  

notperm npgetnode(void) 
{  
    notperm p;  
    p = (notperm) (malloc(sizeof(struct node)));  
    return p;  
}   

int hgetnode() 
{  
    int k;  
    k = i++;  
    return k;  
} 

int agetnode() 
{  
    int k;  
    k = j++;  
    return k;  
} 

void hfreenode(int r) 
{  
    --i;  
}  
void afreenode(int r) 
{  
    --j;  
} 

int addnode(int *pgraph, int x) 
{  
    int p;  
    p = hgetnode();  
    hnode[p].info = x;  
    hnode[p].epoint = -1;  
    hnode[p].tpoint = -1;  
    hnode[p].next = *pgraph;  
    *pgraph = p;  
    return p;  
} 

void joinwt(int p, int q, int wt) 
{  
    int r, r2;  
    int t, t2;  
    r2 = -1;  
    r = hnode[p].epoint; 

    while (r >= 0 && anode[r].epoint != q) 
    {  
     r2 = r;  
     r = anode[r].enext;  
    }   
    if (r >= 0) 
    {  
     anode[r].info = wt;  
    }   
    if (r < 0) 
    {  
     r = agetnode();  
     anode[r].epoint = q;  
     anode[r].tpoint = -1;  
     anode[r].enext = -1;  
     anode[r].tnext = -1;  
     anode[r].info = wt;  
     (r2 < 0) ? (hnode[p].epoint = r) : (anode[r2].enext = r); 
    } 

    /*updation to q */ 

    t = hnode[q].tpoint;  
    t2 = -1;  
    while (t >= 0 && anode[r].tpoint != p) 
    {  
     t2 = t;  
     t = anode[t].tnext;  
    }  
    if (t >= 0) 
    {  
     anode[t].info = wt;  
    } 
    if (t < 0) 
    {  
     t = agetnode();  
     anode[t].tpoint = p;  
     anode[t].epoint = -1; 
     anode[t].tnext = -1;  
     anode[t].enext = -1;  
     anode[t].info = wt;  
     (t2 < 0) ? (hnode[q].tpoint = t) : (anode[t2].tnext = t);  
    } 
} 

void insertnperm(notperm * pq, int x, int currep) 
{  
    notperm p;  
    if (*pq == NULL)  
    {  
     *pq = (notperm) (malloc(sizeof(struct node)));  
     (*pq)->info = x;  
     (*pq)->i = currep;  
     (*pq)->next = NULL;  
     return;  
    }  
    p = npgetnode();  
    p->info = x;  
    p->i = currep;  
    p->next = (*pq);  
    *pq = p;  
}  

void order(notperm * pq, int s, int distance[]) 
{  
    notperm p, q, j;  
    q = NULL;  
    p = NULL;  
    p = *pq;  
    while ((p->i/*this statement always ends up with a segmentation fault*/) != s) 
    {  
     q = p;  
     p = p->next;  
    }  
    if (q == NULL)  
     *pq = p->next;  
    else  
     q->next = p->next;  
    q = NULL;  
    for (j = *pq; distance[(j->i)] < distance[(p->i)]; j = j->next)  
    q = j;  
    if (q == NULL) {  
     p->next = *pq;  
     *pq = p;  
    }  
    else  
    {  
     p->next = q->next;  
     q->next = p;  
    }  
} 

void chngstozero(notperm * pq, int s) 
{  
    notperm p;  
    for (p = *pq; (p->i) != s; p = p->next);  
     p->info = 0;  
    }  
    int checkstat(notperm * pq, int s) 
    {  
     notperm p; 
     for (p = *pq; (p->i) != s; p = p->next);  
     return (p->info);  
    }   
    int newminel(notperm * pq) 
    {  
     notperm p;  
     for (p = *pq; (p->info) != 1; p = p->next);  
     p->info = 0;  
     return (p->i);  
    }   

    void Dijkstra(int graph, int s, int t, int *pd) 
    {  
     unsigned int distance[MAXNODES], precede[MAXNODES];  
     int current, i, k, dc, currep;  
     unsigned int smalldist, newdist;  
     notperm p;  
     notperm nperm;  
     nperm = NULL; 
     for (i = graph; hnode[i].next >= 0;) 
     {  
      insertnperm(&nperm, 1, i);  
      distance[i] = 500000;  
      i = hnode[i].next; 
     }  
     distance[s] =0; 
     order(&nperm, s, distance);  
     chngstozero(&nperm, s); 
     current = s; 
     while (current != t) 
     { 
      smalldist = 500000; 
      dc = distance[current]; 
      for (currep = hnode[current].epoint; anode[currep].enext >= 0; currep = anode[currep].enext) 
      { 
       if (checkstat(&nperm, currep)) 
       { 
        newdist = dc + anode[currep].info; 
        if (newdist < distance[currep]) 
        { 
         distance[currep] = newdist; 
         order(&nperm, currep, distance); 
         precede[currep] = current; 
        } 
       } 
      } 

main() 
{ 
    int graph,a,b,c,d,e,wt,r,t,pd; 
    graph=-1; 
    int wt1=5; 
    int wt2=3; 
    int wt3=6; 
    int wt4=2; 
    int wt5=7; 
    a=addnode(&graph,4); 
    b=addnode(&graph,6); 
    c=addnode(&graph,8); 
    d=addnode(&graph,9); 
    e=addnode(&graph,5); 
    joinwt(a,b,wt); 
    joinwt(a,c,wt1); 
    joinwt(a,d,wt2); 
    joinwt(a,e,wt3); 
    joinwt(d,b,wt4); 
    joinwt(c,e,wt5); 
    Dijkstra(graph,a,e,&pd); 
    printf("%d",pd); 

    getch(); 
    return 0; 
} 
+0

你能解決格式嗎?縮進代碼並擺脫所有的雙間距? – Barmar

+0

@JoachimPileborg謝謝,但我已經知道了。那我怎麼知道分段錯誤在哪裏。 – 10111

回答

0
while ((p->i/*this statement always ends up with a segmentation fault*/) != s) { 
    q = p; 
    p = p->next; 
} 

它看起來像你正在運行關閉列表的末尾指向p。最後一個元素將有NULLp->next;然後設置p = p->next並嘗試解除引用指針,導致分段錯誤。

解決的辦法是在檢查p->i之前檢查NULLp