HDU1372,BFS象棋马走日

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDU1372,BFS象棋马走日相关的知识,希望对你有一定的参考价值。

简单的对于bfs的运用,但是还是写的太慢了写了TMD的1H,主要是不熟悉,以后慢慢熟悉就好了,模型基本已经能建立了,主要出现bug是在方向数组的运用上面,一定要记得是从0开始的,而不是从1开始的,导致错误。

#include<cstdio>
#include<cstdlib>
#include<iostream>

using namespace std;
/*bfs*/

//方向数组
int nextx[8][2]={
    {-1,-2},
    {-2,-1},
    {1,2},
    {2,1},
    {1,-2},
    {2,-1},
    {-1,2},
    {-2,1}
};

/*
1 0 3 0 5 6 7 8 9
0 2 3 4 0 6 7 8 9 
1 2 0 4 5 6 7 8 9
0 2 3 4 0 6 7 8 9
1 0 3 0 5 6 7 8 9 
1 2 3 4 5 6 7 8 9

*/

struct que
{
    int x;//当前的A;
    int y;//当前的B;
    int step;//走过的步数
};

int main()
{
    int i;//循环变量
    int y1,y2;
    int x1,x2;
    int nowX,nowY;
    char ch[10];
    
    int head=1;
    int tail=1;

    struct que ques[100];

    while (gets(ch) != NULL)
    {
        //将输入的数转换成对应数字
        x1 = ch[0]-a+1;
        y1 = ch[1]-0;
        x2 = ch[3]-a+1;
        y2 = ch[4]-0;

        //初始化地图,用于记录走过的点
        int maps[10][10]={0};

        head=1;
        tail=1;

        maps[x1][y1] = 1;//记录初始点
        ques[head].x = x1;
        ques[head].y = y1;
        ques[head].step = 0;

        tail++;
        nowX=x1;
        nowY=y1;
        if(nowX == x2 && nowY == y2)
        {
            goto f1;
        }
        
        while (head<tail)
        {
            for (i = 0; i < 8; i++)
            {
                nowX = ques[head].x + nextx[i][0];
                nowY = ques[head].y + nextx[i][1];

                if(nowX<=0 || nowX>8 || nowY<=0 || nowY>8)
                    continue;
                if(maps[nowX][nowY] == 0)
                {
                    maps[nowX][nowY] = 1;
                    ques[tail].x = nowX;
                    ques[tail].y = nowY;
                    ques[tail].step = ques[head].step + 1;
                    tail++;
                }

                if(nowX == x2 && nowY == y2)
                {
                    goto f1;
                }
            }
            head++;
        }

        f1:printf("To get from %c%d to %c%d takes %d knight moves.\n",ch[0],y1,ch[3],y2,ques[tail-1].step);
    }
    return 0;
}

/*

Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

Sample Output

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.


*/

以上是关于HDU1372,BFS象棋马走日的主要内容,如果未能解决你的问题,请参考以下文章

HDU - 1372 Knight Moves(bfs入门)

骑士巡游的问题简述如下:在国际象棋盘上某一位置放置一个马的棋子,然后采用象棋中“马走日字”规则

马走日问题(Java版)

马走日

002:马走日

问题 D: 马走日