Shortest path of the king(比赛2.A)

最新推荐文章于 2022-01-24 19:24:07 发布

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最新推荐文章于 2022-01-24 19:24:07 发布

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本文探讨了在棋盘上，国王从一个位置移动到另一个特定位置的最短路径问题，详细介绍了使用广度优先搜索（BFS）解决该问题的方法，并提供了正确的代码实现。

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1.题目：

Home	Problems	Status	Contest	[acgirl]	Logout
NEFU要崛起——第2场
3:00:00
OverviewProblemStatusRankDiscuss

A
 
B
 
C
 
D
A - Shortest path of the king
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
 
Practice
 
CodeForces 3A
Description
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.


In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input
The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.

Output
In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.

L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Sample Input
Input
a8
h1
Output
7
RD
RD
RD
RD
RD
RD
RD
FAQ | About Virtual Judge | Forum | Discuss | Open Source Project
All Copyright Reserved ©2010-2012 HUST ACM/ICPC TEAM 
Anything about the OJ, please ask in the forum, or contact author:Isun
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2.题意：已知起点，终点，可以上、下、左、右、左上、左下、右上、右下8个方向走。输出最短路径及每步走的方向

3.思路：bfs

4.一直wrong,待改正

wrong代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;
int sx,sy,ex,ey;
int path[100];
int dir[8][2]={1,1,1,-1,-1,-1,-1,1,0,-1,1,0,-1,0,0,1};
struct node
{
    int x;
    int y;
    int px,py;
    int step;
    int d;
    int visit;
}a[10][10];
int bfs(int n)
{
    memset(a,0,sizeof(a));
    queue<node> q;
    node cur;
    a[sx][sy].x=sx;
    a[sx][sy].y=sy;
    a[sx][sy].step=0;
    a[sx][sy].visit=1;
    q.push(a[sx][sy]);
    int aa=1<<29;
    while(!q.empty())
    {
        cur=q.front();//赋值
        q.pop();
        if(cur.x==ex&&cur.y==ey)
        {
            //printf("cur=%d\n",cur.step);
           aa=cur.step;
            break;
        }
        for(int i=0;i<8;i++)
        {
            int tx=cur.x+dir[i][0];
            int ty=cur.y+dir[i][1];
            if(tx>0&&tx<=n&&ty>0&&ty<=n&&a[tx][ty].visit==0)
            {
                a[tx][ty].visit=1;
                a[tx][ty].px=cur.x;
                a[tx][ty].py=cur.y;
                a[tx][ty].x=tx;
                a[tx][ty].y=ty;
                a[tx][ty].step=cur.step+1;
                a[tx][ty].d=i;
                q.push(a[tx][ty]);
            }
        }
    }
    return  aa;
}
int main()
{
    char s,e;
    int n=8;
    scanf("%c%d",&s,&sy);
    getchar();
    sx=(s-'a')+1;
    scanf("%c%d",&e,&ey);
    getchar();
    ex=(e-'a')+1;
    int w=bfs(n);
    printf("%d\n",w);
    if(w==0)
    return 0;
    int tx=a[ex][ey].px,ty=a[ex][ey].py;
    int i=0;
    path[i++]=a[ex][ey].d;
    while(tx!=sx&&ty!=sy)
    {
        path[i++]=a[tx][ty].d;
        tx=a[tx][ty].px;
        ty=a[tx][ty].py;
    }
    for(int j=i-1;j>=0;j--)
    {
        switch(path[j])
        {
            case 0:
            printf("RU\n");
            break;
            case 1:
            printf("RD\n");
            break;
            case 2:
            printf("LD\n");
            break;
            case 3:
            printf("LU\n");
            break;
            case 4:
            printf("D\n");
            break;
            case 5:
            printf("R\n");
            break;
            case 6:
            printf("L\n");
            break;
            case 7:
            printf("RU\n");
            break;
        }
    }
    return 0;
}

4.accept代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int move[8][2]={{0,-1},{0,1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
//             左 L    右  R 下  D  上 U    右下 RD  左下  LD  右上RU    左上LU
char mo[8][3]={'L','R','D','U','RD','LD','RU','LU'};
int ex,ey,sx,sy;
struct node{
    int x,y;
    int vis;
    int depth;
    int shx,shy;
    int m;
}pan[10][10];
int bfs()
{
    int i,j,k;
    queue<node> q;
    q.push(pan[sx][sy]);
    if(sx==ex&&sy==ey)
    return 0;
    pan[sx][sy].vis=1;
    while(!q.empty())
    {
        node p;
        int tx,ty,yx,yy;
        p=q.front();
        q.pop();

            for(i=0;i<8;i++)
            {
				node t;
				t=p;
                tx=p.x+move[i][0];
                ty=p.y+move[i][1];
                if(tx<0||tx>=8||ty<0||ty>=8)
                    continue;
                if(pan[tx][ty].vis==0)
                {
                    pan[tx][ty].shx=p.x;
					pan[tx][ty].shy=p.y;
                    pan[tx][ty].depth=p.depth+1;
                    q.push(pan[tx][ty]);
                    pan[tx][ty].vis=1;
                    pan[tx][ty].m=i;

                    if(tx==ex&&ty==ey)
                        return pan[tx][ty].depth;
                }
            }


    }
    return 0;
}
void init()
{
    int i,j;
    for(i=0;i<8;i++)
    {
        for(j=0;j<8;j++)
        {
            pan[i][j].x=i;
            pan[i][j].y=j;
        }
    }
}
int main()
{
    int i,j,k,l,m,n;
    memset(pan,0,sizeof(pan));
    init();
    char z,v;
    scanf("%c%c",&z,&v);//先y(字母)后x
    sy=z-'a';
	sx=8-v+48;
	getchar();
    scanf("%c%c",&z,&v);
    ey=z-'a';
	ex=8-v+48;
	m=bfs();
    printf("%d\n",m);
    if(m==0)
    return 0;
    int  path[100];
    memset(path,0,sizeof(path));
    j=0;
    path[j++]=pan[ex][ey].m;
	int qx,qy;
	qx=pan[ex][ey].shx;
	qy=pan[ex][ey].shy;
    while(qx!=sx||qy!=sy)
    {
		int tx,ty;
		tx=qx;ty=qy;
		qx=pan[tx][ty].shx;
		qy=pan[tx][ty].shy;
        path[j++]=pan[tx][ty].m;
        //printf("%d\n",p->m);
    }
    for(i=j-1;i>=0;i--)
	{
		int k=path[i];
		if(k==0)
		printf("L\n");
		else if(k==1)
		printf("R\n");
		else if(k==2)
		printf("D\n");
		else if(k==3)
		printf("U\n");
		else if(k==4)
		printf("RD\n");
		else if(k==5)
		printf("LD\n");
		else if(k==6)
		printf("RU\n");
		else if(k==7)
		printf("LU\n");
	}

return 0;
}