toj 1702 A Knight's Journey
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?
ProblemFind a path such that the knight visits every square once. The knight can start and end on any square of the board.
InputThe 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, . . .
OutputThe 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.
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
Source: TUD Programming Contest 2005
Problem ID in problemset: 1702
//#include <stdio.h>
int t;
int main()
{
scanf("%d",&t);
int p,q,zz=1;
while(t--)
{
scanf("%d%d",&p,&q);
printf("Scenario #%d:\n",zz++);
if(p==1&&q==1)
printf("A1\n");
else if(p==3&&q==4)
printf("A1C2A3B1D2B3C1A2C3D1B2D3\n");
else
if(p==3&&q==7)
printf("A1B3D2F1G3E2G1F3E1G2E3C2A3B1C3A2C1D3B2D1F2\n");
else if(p==3&&q==8)
printf("A1B3C1A2C3D1B2D3E1G2E3C2A3B1D2F1H2F3G1E2G3H1F2H3\n");
else
if(p==4&&q==3)
printf("A1B3C1A2B4C2A3B1C3A4B2C4\n");
else
if(p==4&&q==5)
printf("A1B3C1A2B4D3E1C2D4E2C3A4B2D1E3C4A3B1D2E4\n");
else
if(p==4&&q==6)
printf("A1B3C1A2B4C2D4E2F4D3E1F3D2B1A3C4B2A4C3E4F2D1E3F1\n");
else
if(p==5&&q==4)
printf("A1B3A5C4D2B1A3B5D4C2B4A2C1D3C5A4B2D1C3D5\n");
else
if(p==5&&q==5)
printf("A1B3A5C4A3B1D2E4C5A4B2D1C3B5D4E2C1A2B4D5E3C2E1D3E5\n");
else
if(p==6&&q==4)
printf("A1B3A5C6D4B5D6C4D2B1A3C2B4A2C1D3B2D1C3D5B6A4C5A6\n");
else
if(p==7&&q==3)
printf("A1B3C1A2C3B1A3C2B4A6C7B5A7C6A5B7C5A4B2C4B6\n");
else
if(p==8&&q==3)
printf("A1B3C1A2B4C2A3B1C3A4B2C4A5B7C5A6B8C6A7B5C7A8B6C8\n");
else
printf("impossible\n");
printf("\n");
}
return 0;
}
//比赛的时候自己错了一点,赛后交了是对的,唉,深搜一定要细心啊!!!!
//不打表也是对的
#include <iostream>
#include <stack>
#define MAX 27
using namespace std;
bool used[MAX][MAX],mark;
int sum,p,q,t;
int a[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int num;
typedef struct node
{
short num;
char ch;
node(){};
node(short nn,char ccc)
{
num=nn;
ch=ccc;
}
}Point;
stack<Point>S,RS;
Point temp;
void Init()
{
scanf("%d%d",&p,&q);
int i,j;
for(i=0;i<q;i++)
for(j=0;j<p;j++)
used[i][j]=false;
}
bool Bound(int x,int y)
{
if(x>=0&&y>=0&&x<p&&y<q)
return true;
else
return false;
}
void DFS(int x,int y,int sum) //x是列,y是行
{
int i,j,tx,ty;
if(sum==p*q)
{
mark=true;
return;
}
for(i=0;i<8;i++)
{
tx=x+a[i][1]; //tx是列
ty=y+a[i][0]; //ty是行
if(!Bound(ty,tx)||used[tx][ty])
{
continue;
}
used[tx][ty]=true;
S.push(node(ty,(char)('A'+tx)));
sum+=1;
DFS(tx,ty,sum);
if(!mark)
S.pop();
if(mark)
return ;
used[tx][ty]=false;
sum-=1;
}
}
int main()
{
int i,j;
int zz=1;
scanf("%d",&t);
while(t--)
{
Init();
printf("Scenario #%d:\n",zz++);
mark=false;
num=0;
for(i=0;i<q;i++)
{
if(mark)
break;
for(j=0;j<p;j++)
{
used[i][j]=true;
while(!S.empty())
S.pop();
sum=1;
temp.ch=(char)('A'+i);
temp.num=j;
S.push(temp);
DFS(i,j,sum); //i是列,j是行
if(mark)
break;
used[i][j]=false;
}
}
if(!mark)
printf("impossible\n\n");
else
{
while(!S.empty())
{
temp=S.top();
S.pop();
RS.push(temp);
}
while(!RS.empty())
{
temp=RS.top();
RS.pop();
printf("%c%d",temp.ch,temp.num+1);
}
printf("\n\n");
}
}
return 0;
}
