全国交通咨询模拟报告

全国交通咨询模拟

r->adjvex)).stata[0].begintime[0],(*(*(arcs+q->adjvex)+r->adjvex)).stata[0].begintime[1],G.vertices[q->adjvex].cityname,G.vertices[r->adjvex].cityname);

q=r;

r=r->next; n++;

}

printf(\最少中转次数是%d次\\n\\n\

for(v=0;v

{

q=p[v].next;

while(q!=NULL)

{

s=q;

q=q->next; free(s);

}

for(v=0;v

q=p[v].next; }

}

t=t->nextarc;

}

EnterQueue(&Q,w);

} free(p); return;

}

p[v].next=NULL;

while(q!=NULL)

29

全国交通咨询模拟

{

s=q;

q=q->next; free(s);

}

p[v].next=NULL;

}

free(p);

if(k==1)

printf(\不存在列车车次从%s到

%s\\n\\n\

else

printf(\不存在飞机航班从%s到

%s\\n\\n\

}

void MinExpenditure(infolist arcs,float *expenditure,int *route) {

int i;

*expenditure=arcs.stata[0].expenditure; if(*expenditure

*route=0;

else

*route=-1;

for(i=1;i<=arcs.last;i++)

if(arcs.stata[i].expenditure<*expenditure) {

*expenditure=arcs.stata[i].expenditure;

*route=i; }

30

}

全国交通咨询模拟

4.4.求城市v0，v1之间的最少费用程序：

void ExpenditureDispose(int k,infolist (*arcs)[MAX_VERTEX_NUM],ALGraph G,int v0,int v1,float *M,int *final)

{

int v=-1,w,i,route;

float m,expenditure; Node *p,*q,*r,*s;

p=(Node *)malloc(G.vexnum*sizeof(Node)); for(v=0;v

{

*(final+v)=False;

MinExpenditure(*(*(arcs+v0)+v),M+v,&route); p[v].next=NULL; if(*(M+v)

{

q=(Node *)malloc(sizeof(Node));

s=(Node *)malloc(sizeof(Node)); q->adjvex=v0; s->adjvex=v; s->route=route; p[v].next=q; q->next=s; s->next=NULL;

}

*(M+v0)=0;

}

*(final+v0)=True; for(i=1;i

{

31

全国交通咨询模拟

m=INFINITY;

v=-1;

for(w=0;w

if(*(final+w)==False)

if(*(M+w)

v=w;

m=*(M+w);

}

if(v==v1) {

q=p[v].next;

r=q->next;

printf(\旅行路线是:\\n\ while(r!=NULL)

{

if(k==1)

printf(\乘坐No.%d列车车次在%d:%d从%s到%

s\\n\jvex)+r->adjvex)).stata[r->route].begintime[0],(*(*(arcs+q->adjvex)+r->adjvex)).stata[r->route].begintime[1],G.vertices[q->adjvex].cityname,G.vertices[r->adjvex].cityname);

else

printf(\乘坐No.%d飞机航班在%d:%d从%s到%s\\n\djvex)+r->adjvex)).stata[r->route].begintime[0],(*(*(arcs+q->adjvex)+r->adjvex)).stata[r->route].begintime[1],G.vertices[q->adjvex].cityname,G.vertices[r->adjvex].cityname);

q=r; r=r->next;

32