1.代码:
#include <stdio.h>
#include <malloc.h>
#define NoEdge 1000
struct MinHeapNode
{
int lcost; //子树费用的下界
int cc; //当前费用
int rcost; //x[s:n-1]中顶点最小出边费用和
int s; //根节点到当前节点的路径为x[0:s]
int *x; //需要进一步搜索的顶点是//x[s+1:n-1]
struct MinHeapNode *next;
};
int n; //图G的顶点数
int **a; //图G的邻接矩阵
//int NoEdge; //图G的无边标记
int cc; //当前费用
int bestc; //当前最小费用
MinHeapNode* head = 0; /*堆头*/
MinHeapNode* lq = 0; /*堆第一个元素*/
MinHeapNode* fq = 0; /*堆最后一个元素*/
int DeleteMin(MinHeapNode*&E)
{
MinHeapNode* tmp = NULL;
tmp = fq;
// w = fq->weight ;
E = fq;
if(E == NULL)
return 0;
head->next = fq->next; /*一定不能丢了链表头*/
fq = fq->next;
// free(tmp) ;
return 0;
}
int Insert(MinHeapNode* hn)
{
if(head->next == NULL)
{
head->next = hn; //将元素放入链表中
fq = lq = head->next; //一定要使元素放到链中
}else
{
MinHeapNode *tmp = NULL;
tmp = fq;
if(tmp->cc > hn->cc)
{
hn->next = tmp;
head->next = hn;
fq = head->next; /*链表只有一个元素的情况*/
}else
{
for(; tmp != NULL;)
{
if(tmp->next != NULL && tmp->cc > hn->cc)
{
hn->next = tmp->next;
tmp->next = hn;
break;
}
tmp = tmp->next;
}
}
if(tmp == NULL)
{
lq->next = hn;
lq = lq->next;
}
}
return 0;
}
int BBTSP(int v[])
{//解旅行售货员问题的优先队列式分支限界法
/*初始化最优队列的头结点*/
head = (MinHeapNode*)malloc(sizeof(MinHeapNode));
head->cc = 0;
head->x = 0;
head->lcost = 0;
head->next = NULL;
head->rcost = 0;
head->s = 0;
int *MinOut = new int[n + 1]; /*定义定点i的最小出边费用*/
//计算MinOut[i]=顶点i的最小出边费用
int MinSum = 0;//最小出边费用总合
for(int i = 1; i <= n; i++)
{
int Min = NoEdge; /*定义当前最小值*/
for(int j = 1; j <= n; j++)
if(a[i][j] != NoEdge && /*当定点i,j之间存在回路时*/
(a[i][j] < Min || Min == NoEdge)) /*当顶点i,j之间的距离小于Min*/
Min = a[i][j]; /*更新当前最小值*/
if(Min == NoEdge)
return NoEdge;//无回路
MinOut[i] = Min;
//printf("%d\n",MinOut[i]);/*顶点i的最小出边费用*/
MinSum += Min;
// printf("%d\n",MinSum); /*最小出边费用的总和*/
}
MinHeapNode *E = 0;
E = (MinHeapNode*)malloc(sizeof(MinHeapNode));
E->x = new int[n];
// E.x=new int[n];
for(int i = 0; i < n; i++)
E->x[i] = i + 1;
E->s = 0;
E->cc = 0;
E->rcost = MinSum;
E->next = 0; //初始化当前扩展节点
int bestc = NoEdge; /*记录当前最小值*/
//搜索排列空间树
while(E->s < n - 1)
{//非叶结点
if(E->s == n - 2)
{//当前扩展结点是叶结点的父结点
/*
首先考虑s=n-2的情形,此时当前扩展结点是排列树中某个叶结点的父结点。如果该叶结点相应一条可行回路
且费用小于当前最小费用,则将该叶结点插入到优先队列中,否则舍去该叶结点
*/
if(a[E->x[n - 2]][E->x[n - 1]] != NoEdge && /*当前要扩展和叶节点有边存在*/
a[E->x[n - 1]][1] != NoEdge && /*当前页节点有回路*/
(E->cc + a[E->x[n - 2]][E->x[n - 1]] + a[E->x[n - 1]][1] < bestc /*该节点相应费用小于最小费用*/
|| bestc == NoEdge))
{
bestc = E->cc + a[E->x[n - 2]][E->x[n - 1]] + a[E->x[n - 1]][1]; /*更新当前最新费用*/
E->cc = bestc;
E->lcost = bestc;
E->s++;
E->next = NULL;
Insert(E); /*将该页节点插入到优先队列中*/
}else
free(E->x);//该页节点不满足条件舍弃扩展结点
}else
{/*产生当前扩展结点的儿子结点
当s<n-2时,算法依次产生当前扩展结点的所有儿子结点。由于当前扩展结点所相应的路径是x[0:s],
其可行儿子结点是从剩余顶点x[s+1:n-1]中选取的顶点x[i],且(x[s],x[i])是所给有向图G中的一条边。
对于当前扩展结点的每一个可行儿子结点,计算出其前缀(x[0:s],x[i])的费用cc和相应的下界lcost。
当lcost<bestc时,将这个可行儿子结点插入到活结点优先队列中。*/
for(int i = E->s + 1; i < n; i++)
if(a[E->x[E->s]][E->x[i]] != NoEdge)
{ /*当前扩展节点到其他节点有边存在*/
//可行儿子结点
int cc = E->cc + a[E->x[E->s]][E->x[i]]; /*加上节点i后当前节点路径*/
int rcost = E->rcost - MinOut[E->x[E->s]]; /*剩余节点的和*/
int b = cc + rcost; //下界
if(b < bestc || bestc == NoEdge)
{//子树可能含最优解,结点插入最小堆
MinHeapNode * N;
N = (MinHeapNode*)malloc(sizeof(MinHeapNode));
N->x = new int[n];
for(int j = 0; j < n; j++)
N->x[j] = E->x[j];
N->x[E->s + 1] = E->x[i];
N->x[i] = E->x[E->s + 1];/*添加当前路径*/
N->cc = cc; /*更新当前路径距离*/
N->s = E->s + 1; /*更新当前节点*/
N->lcost = b; /*更新当前下界*/
N->rcost = rcost;
N->next = NULL;
Insert(N); /*将这个可行儿子结点插入到活结点优先队列中*/
}
}
free(E->x);
}//完成结点扩展
DeleteMin(E);//取下一扩展结点
if(E == NULL)
break; //堆已空
}
if(bestc == NoEdge)
return NoEdge;//无回路
for(int i = 0; i < n; i++)
v[i + 1] = E->x[i];//将最优解复制到v[1:n]
while(true)
{//释放最小堆中所有结点
free(E->x);
DeleteMin(E);
if(E == NULL)
break;
}
return bestc;
}
int main()
{
n = 0;
int i = 0;
//FILE *in, *out;
//in = fopen("input.txt", "r");
//out = fopen("output.txt", "w");
//if(in == NULL || out == NULL)
//{
// printf("没有输入输出文件\n");
// return 1;
//}
//fscanf(in, "%d", &n);
n=5;
a = (int**)malloc(sizeof(int*) * (n + 1));
for(i = 1; i <= n; i++)
{
a[i] = (int*)malloc(sizeof(int) * (n + 1));
}
// for(i = 1; i <= n; i++)
// for(int j = 1; j <= n; j++)
// //fscanf(in, "%d", &a[i][j]);
// a[i][j]=1;
a[1][1]=0;
a[1][2]=5;
a[1][3]=8;
a[1][4]=5;
a[1][5]=4;
a[2][1]=5;
a[2][2]=0;
a[2][3]=5;
a[2][4]=6;
a[2][5]=3;
a[3][1]=8;
a[3][2]=5;
a[3][3]=0;
a[3][4]=5;
a[3][5]=4;
a[4][1]=5;
a[4][2]=6;
a[4][3]=6;
a[4][4]=0;
a[4][5]=3;
a[5][1]=4;
a[5][2]=3;
a[5][3]=4;
a[5][4]=3;
a[5][5]=0;
// prev = (int*)malloc(sizeof(int)*(n+1)) ;
int*v = (int*)malloc(sizeof(int) * (n + 1));// MaxLoading(w , c , n) ;
for(i = 1; i <= n; i++)
v[i] = 0;
bestc = BBTSP(v);
printf("\n");
for(i = 1; i <= n; i++)
fprintf(stdout, "%d\t", v[i]);
fprintf(stdout, "\n");
fprintf(stdout, "%d\n", bestc);
return 0;
}
2.输出结果
1->2->5->3->4->1
距离:22