马的Hamilton周游路线问题

来源：互联网发布：抓包工具二次开发源码编辑：程序博客网时间：2024/06/02 15:19
#include<iostream>#include <iomanip>#include <queue>using namespace std;//在某一格子的八种走法int fx[]= {2,1,-1,-2,-2,-1,1,2};int fy[]= {-1,-2,-2,-1,1,2,2,1};typedef struct{    int x,y; //坐标    int number; //序号} Point; //棋盘中的格子//马周游的棋盘，注意使用的时候是从下表为1开始Point board[10000][10000];int n; //棋盘大小int step =1; //序号//输出结果void outputResult(int n){    for(int i=1; i<=n; i++)    {        cout<<endl<<endl;        for(int j=1; j<=n; j++) {            cout<<setw(3)<<board[i][j].number<<" ";        } cout<<endl<<endl;    }}bool check(int x,int y) {    if(x<1 || y<1 || x>n || y>n || board[x][y].number != 0)        return false;    return true;}//下一位置有多少种走法int nextPosHasSteps(int x, int y){    int steps = 0;    for (int i = 0; i < 8; ++i)    {        if (check(x + fx[i], y + fy[i]))            steps++;    }    return steps;}//非递归的走法void horseRun(Point point){    queue<Point> pointQueue;    pointQueue.push(point);    Point temp;    while(!pointQueue.empty())    {        temp = pointQueue.front();        pointQueue.pop();        board[temp.x][temp.y].number = step++;        int minStep = 8;        int flag = 0;        for(int i=0; i<8; i++) //出下一位置走法最少的进入对列        {            int x=temp.x + fx[i];            int y=temp.y + fy[i];            if(check(x,y))            {                if(nextPosHasSteps(x,y) <= minStep)                {                    minStep = nextPosHasSteps(x,y);                    Point t;                    t.x = x;                    t.y = y;                    if(flag) pointQueue.pop();                    pointQueue.push(t);                    flag = 1;                }            }        }    }}int main(){    cout<<"输入棋盘大小n:";    cin>>n;    Point startPoint;    cout<<"输入马周游起始位置x(1~n),y(1~n):";    cin>>startPoint.x>>startPoint.y;    horseRun(startPoint);    //输出结果    outputResult(n);    return 0;}
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