A Knight's Journey--DFS

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 40883 Accepted: 13906

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题目链接：http://poj.org/problem?id=2488

题目大意：

这个题的意思是说输入两个数代表棋盘的大小，然后骑士固定从左上角开始走，问能否使骑士走过整张棋盘，如果可以，输出路径，注意，按字典序遍历。

按照此顺序遍历。

我必须吐槽一下这个题，我和柏皓找bug找了整整一个上午，就是一个很水的dfs，但是，最后找到的bug结果是，我用结构体存的路径，他用的字符数组，然后就。。。我真服气了。在此放两个代码，一个是我的最开始的版本，一个是2.0版本。一样的思路。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int h;int vis[200][200];int x,y;struct node{    int a;    int b;}xin[10000];int top;int kkk;int flag;char s[128],a[128];void dfs(int sx,int sy,int step){    //printf("%d %d\n",step,h);    if(step==h)    {         s[top]=0;        strcpy(a,s);        flag=1;    }    if(flag)    {        return ;    }    if(sx-1>=1&&sy-2>=1&&!vis[sx-1][sy-2])//字典序8个方向    {        s[top++]=sy-2+'A'-1;            s[top++]=sx-1+'0';        vis[sx-1][sy-2]=1;        dfs(sx-1,sy-2,step+1);        vis[sx-1][sy-2]=0;        top-=2;    }    if(sx+1<=x&&sy-2>=1&&!vis[sx+1][sy-2])//一样的套路    {        s[top++]=sy-2+'A'-1;            s[top++]=sx+1+'0';        vis[sx+1][sy-2]=1;        dfs(sx+1,sy-2,step+1);        vis[sx+1][sy-2]=0;        top-=2;    }    if(sx-2>=1&&sy-1>=1&&!vis[sx-2][sy-1])    {        s[top++]=sy-1+'A'-1;            s[top++]=sx-2+'0';        vis[sx-2][sy-1]=1;        dfs(sx-2,sy-1,step+1);        vis[sx-2][sy-1]=0;        top-=2;    }    if(sx+2<=x&&sy-1>=1&&!vis[sx+2][sy-1])    {        s[top++]=sy-1+'A'-1;            s[top++]=sx+2+'0';        vis[sx+2][sy-1]=1;        dfs(sx+2,sy-1,step+1);        vis[sx+2][sy-1]=0;      top-=2;    }    if(sx-2>=1&&sy+1<=y&&!vis[sx-2][sy+1])    {        s[top++]=sy+1+'A'-1;            s[top++]=sx-2+'0';        vis[sx-2][sy+1]=1;        dfs(sx-2,sy+1,step+1);        vis[sx-2][sy+1]=0;      top-=2;    }    if(sx+2<=x&&sy+1<=y&&!vis[sx+2][sy+1])    {       s[top++]=sy+1+'A'-1;            s[top++]=sx+2+'0';        vis[sx+2][sy+1]=1;        dfs(sx+2,sy+1,step+1);        vis[sx+2][sy+1]=0;        top-=2;    }    if(sx-1>=1&&sy+2<=y&&!vis[sx-1][sy+2])    {       s[top++]=sy+2+'A'-1;            s[top++]=sx-1+'0';        vis[sx-1][sy+2]=1;        dfs(sx-1,sy+2,step+1);        vis[sx-1][sy+2]=0;        top-=2;    }    if(sx+1<=x&&sy+2<=y&&!vis[sx+1][sy+2])    {        s[top++]=sy+2+'A'-1;            s[top++]=sx+1+'0';        vis[sx+1][sy+2]=1;        dfs(sx+1,sy+2,step+1);        vis[sx+1][sy+2]=0;        top-=2;    }}int main(){    int t;    scanf("%d",&t);    int kk=0;    while(t--)    {        flag=0;        kk++;        memset(vis,0,sizeof(vis));        top=2;        scanf("%d%d",&x,&y);        h=x*y;        s[0]='A',s[1]='1';        xin[0].a=1;        xin[0].b=1;        vis[1][1]=1;        printf("Scenario #%d:\n",kk);        dfs(1,1,1);        if(flag)        {            printf("%s\n\n",a);        }        else            printf("impossible\n\n");    }    return 0;}

结构体太坑了，我到现在为止不知道为什么结构体会出错。

2.0版本代码

#include <cstdio>#include <cstring>using namespace std;struct node{    int a,b;} xin[30];char s[128],a[128];int top,kkk,flag,x,y,vis[30][30],dy[8]= {-2,-2,-1,-1,1,1,2,2},dx[8]= {-1,1,-2,2,-2,2,-1,1};//8个方向void dfs(int sx,int sy,int step){    if(step==x*y)    {        kkk=top;        flag=1;    }    if(flag)        return;    int i,l,r;     for (i=0; i<8; i++)    {        l=sx+dx[i];        r=sy+dy[i];        if (l>0&&l<=x&&r>0&&r<=y&&!vis[l][r])        {            vis[l][r]=1;            xin[top].a=r;            xin[top++].b=l;            dfs(l,r,step+1);            top--;            vis[l][r]=0;        }    }}int main(){    int t,kk=0;    scanf("%d",&t);    while(t--)    {        flag=0;        memset(vis,0,sizeof(vis));        scanf("%d%d",&x,&y);        top=2;        vis[1][1]=1;        s[0]='A',s[1]='1';        printf("Scenario #%d:\n",++kk);        dfs1(1,1,1);        if(flag)            printf("%s\n\n",a);        else            printf("impossible\n\n");       // printf("\n\n");    }    return 0;}

注意此题输出时每一组多输出一行空格。