POJ 2488 A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 23108 Accepted: 7819

Description

 

Background 

The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey 

around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans? 

 

Problem 

Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

 

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

 

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number. 

If no such path exist, you should output impossible on a single line.

Sample Input

 

3

1 1

2 3

4 3

Sample Output

 

Scenario #1:

A1

 

Scenario #2:

impossible

 

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

Source

 

TUD Programming Contest 2005, Darmstadt, Germany

读懂了题目这题就会做了，以前的马在跳的时候都是在线上，这个题是这方格内还真有些不适应，转化成在线上就好

[cpp]  

#include <stdio.h>  

#include <string.h>  

int status[100][100],s;  

int vex[]={-2,-2,2,2,-1,-1,1,1};  

int vey[]={-1,1,-1,1,-2,2,-2,2};  

char path1[1000],prepath1[1000];  

int path2[1000],prepath2[1000],top,n,m,key,k;  

int main()  

{  

    void dfs(int x,int y);  

    int i,j,t,tem;  

    scanf("%d",&t);  

    tem=1;  

    while(t--)  

    {  

        scanf("%d %d",&n,&m);  

        memset(status,0,sizeof(status));  

        if(n==1&&m==1)  

        {  

            printf("Scenario #%d:\n",tem); tem++;  

            printf("A1\n");  

            printf("\n");  

            continue;  

        }  

        for(i=1;i<=n;i++)  

        {  

            for(j=1;j<=m;j++)  

            {  

                top=0;s=1;  

                memset(status,0,sizeof(status));  

                prepath1[top]='A'+j-1;  

                prepath2[top++]=i;  

                status[i][j]=1;  

                key=0; k=0;  

                dfs(i,j);  

                if(k==1)  

                {  

                    break;  

                }  

            }  

            if(j!=m+1)  

            {  

                break;  

            }  

        }  

        printf("Scenario #%d:\n",tem); tem++;  

        if(i!=n+1)  

        {  

            for(i=0;i<=n*m-1;i++)  

            {  

                printf("%c%d",path1[i],path2[i]);  

            }  

            printf("\n");  

        }else  

        {  

            printf("impossible\n");  

        }  

        printf("\n");  

    }  

    return 0;  

}  

void dfs(int x,int y)  

{  

    int i,j,xend,yend,change;  

    for(i=0;i<=7;i++)  

    {  

        xend=x+vex[i]; yend=y+vey[i];  

        if(xend>=1&&xend<=n&¥d>=1&¥d<=m&&!status[xend][yend])  

        {  

            s++;  

            prepath1[top]=yend+'A'-1;  

            prepath2[top++]=xend;  

            status[xend][yend]=1;  

            if(s==n*m)  

            {  

                k=1;  

                if(key==0)  

                {  

                    for(j=0;j<=n*m-1;j++)  

                    {  

                    &

补充：软件开发 , C语言 ,