旅行者问题

基本数据结构省略了,用队列存储状态,即还剩下哪些点没有访问过.然后采取回溯算法进行遍历.

#include "stdafx.h"#include "队列ADT.h"#include "math.h"#include "string.h"#define N 4#define Number (1<<N) -1#define Infinity 1000int AllDistance[N][Number] = {0};int CurrentPath[N][Number];int Find(int start,int Current,int Num,queue NodeCollection,int C[][N]){    int Count = 0;    int Nextnode;    int Distance = Infinity,NextDistance;    if(NodeCollection->size == 0)//如果点集为空    {        CurrentPath[Current][Num] = start;        return C[Current][start];    }    else    {        while(Count++!=NodeCollection->size)//遍历队列中所有点        {             Nextnode = Dequeue(NodeCollection);//从队列中取出一点作为下个目的点            if(AllDistance[Nextnode][Num-(1<<Nextnode)]!=0&&NodeCollection->size!=1)         NextDistance = AllDistance[Nextnode][Num-(1<<Nextnode)];//已经计算过相关数据，不用递归             else//没有相关数据             NextDistance = Find(start,Nextnode,Num-(1<<Nextnode),NodeCollection,C);             Enqueue(NodeCollection,Nextnode);             if(C[Current][Nextnode] + NextDistance < Distance)             {                 Distance = C[Current][Nextnode] + NextDistance;//将最短路径更新                 AllDistance[Current][Num] = Distance;//将相关记录记录                 CurrentPath[Current][Num] = Nextnode; //记录路径             }        }        return Distance;    }}void ReadNodeintoQueue(queue Q){    for(int i = 0;i<=N-1;i++)        Enqueue(Q,i);}int FindMin(int C[][N]){    int start,Minstart;    int Distance,MinDistance = Infinity;    queue Q = (queue)malloc(sizeof(Queue));    Initialize(Q,N);    ReadNodeintoQueue(Q);    for(int i = 0;i<=N-1;i++)//选择一个起始点    {        start = Dequeue(Q);        Distance = Find(start,start,Number-(1<<start),Q,C);        Enqueue(Q,i);        printf("%d: %d\n",i,Distance);        if(Distance < MinDistance)        {            Minstart = start;            MinDistance = Distance;        }    }    return MinDistance;}int _tmain(int argc, _TCHAR* argv[]){    int x = (1<<4);    queue Q = (queue)malloc(sizeof(Queue));    Initialize(Q,5);    int MinDistance;    int C[N][N];    for (int i = 0;i<=N-1;i++)    {        for(int j = 0;j<=N-1;j++)        {            if(i == j)                C[i][j] = 0;            else                C[i][j] = Infinity;        }    }    C[0][1] = 1;    C[0][2] =1;    C[1][0] = 1;    C[1][2] = 2;    C[1][3] = 7;    C[2][0] = 1;    C[2][3] = 9;    C[3][0] = 4;    C[3][2] = 9;    MinDistance = FindMin(C);    return 0;}