题目描述
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?
输入
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, . . .
输出
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.
样例输入
3
1 1
2 3
4 3
样例输出
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
参考代码
#include <stdio.h>
#include <string.h>
#define N 10
int map[N][N];
int str1[100],str2[100];
int dx[8]={-2,-2,-1,-1,1,1,2,2};
//åå
¸é¡ºåº
int dy[8]={-1,1,-2,2,-2,2,-1,1};
void Dfs(int x,int y,int count);
void print();
int p,q;
int flag;
int main()
{
int i,j,m;
int n;
scanf("%d",&n);
for (m=0;m<n;m++)
{
scanf("%d%d",&p,&q);
printf("Scenario #%d:n",m+1);
for(i=0;i<=q;i++)
for(j=0;j<=p;j++)
map[i][j]=0;
flag=0;
map[1][1]=1;
Dfs(1,1,1);
if(flag==0){
printf("impossiblen");
}
else{
for(j=1;j<=p*q;j++){
printf("%c%d",str1[j]+'A'-1,str2[j]);
}
printf("n");
}
printf("n");
}
}
void Dfs(int x,int y,int count)
{
int x1,y1;
int i;
if(p*q==count){
flag=1;
print();
return;
}
for(i=0;i<8;i++)
{
if(flag)
return;
else{
x1=x+dx[i];y1=y+dy[i];
if(x1>=1&&x1<=q&&y1>=1&&y1<=p&&!map[x1][y1])
{
map[x1][y1]=count+1;
Dfs(x1,y1,count+1);
map[x1][y1]=0;
}
}
}
return;
}
void print(){
int i,j;
for(i=1;i<=q;i++)
for(j=1;j<=p;j++){
str1[map[i][j]]=i;
str2[map[i][j]]=j;
}
}
解析
暂无
