pro.rtf

用C语言实现旅行商问题。该算法简单精悍

if(i>>(j-1)&1==0){ //j点不在集合i中
	min=INF;
	for(k=1;k<=CityNum;k++){
	if(i>>(k-1)&1 !=0){  //k点在集合i中
						             Dij[j][i] = Distance[j][k]+Dij[k][i-(1<<(k-1))];//递推公式

就少加了一个括号,结果就错误
涉及到优先级

正确应为if(（i>>(j-1)&1）==0)
D:\TSP\TSP>tsp TSP4.txt
The lowest cost of searched rout is 4
The searched Path is :
  1->4->1->1->1
time :0

D:\TSP\TSP>tsp TSP6.txt
The lowest cost of searched rout is 18
The searched Path is :
  1->3->1->1->1->1->1
time :0



for(j=CityNum-1;j>0;j--){ //逆向求各个结点
  //直接放进数组内有问题
curpos=routine[curpos][k-(1<<(curpos-1))];
printf("->%d",curpos+1);
}

D:\TSP\TSP>tsp TSP6.txt
The lowest cost of searched rout is 175
The searched Path is :
  1->2->4->2->4->2->4
time :79

D:\TSP\TSP>tsp TSP6.txt
The lowest cost of searched rout is 175
The searched Path is :
  1->2->4->2->4->2->4
time :32

D:\TSP\TSP>tsp TSP4.txt
The lowest cost of searched rout is 25
The searched Path is :
  1->3->2->3->2
time :0




The lowest cost of searched rout is 371
The searched Path is :
  1->9->14->8->6->13->5->15->3->4->7->12->2->10->11->1
time :63

D:\TSP\TSP>