搜索进阶题目之A Knight's Journey
时间： 1ms 内存：128M
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.
示例输入：
3
1 1
2 3
4 3示例输出：
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4提示：
参考答案（内存最优[1092]）：
#include <stdio.h>
#include <stdlib.h>
#define MAX 27
int map[MAX][MAX], sx[MAX], sy[MAX];
int dir[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};
int p, q, sign, step;
void dfs(int i, int j)
{
if (sign) return;
int x, y, k;
step++;
sx[step] = i;
sy[step] = j;
if(step == p * q)
{
sign = 1;
return;
}
map[i][j] = 1;
for (k = 0; k < 8; k++)
{
y = j + dir[k][0];
x = i + dir[k][1];
if (map[x][y] == 0 && x > 0 && x <= p && y > 0 && y <= q)
{
dfs(x, y);
step--;
}
}
map[i][j] = 0;
}
int main()
{
int i, j, n, t = 0;
scanf("%d", &n);
while(n--)
{
sign = 0;
step = 0;
t++;
scanf("%d%d", &p, &q);
for (i = 1; i <= p; i++)
{
for(j = 1; j <= q; j++)
{
map[i][j] = 0;
}
}
dfs(1, 1);
printf("Scenario #%d:\n", t);
if (sign)
{
for (i = 1; i <= p * q; i++)
{
printf("%c%d", sy[i] + 64, sx[i]);
}
printf("\n");
}
else
{
printf("impossible\n");
}
if (n != 0) printf("\n");
}
return 0;
}
参考答案（时间最优[4]）：
#include<cstdio>
#include<cstring>
int n,m,nm;
bool map1[15][15],flag;
int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
struct node
{
int x,y;
}record[15][15];
char ans[50]="A1";
void DFS(int x,int y,int step)
{
if(step==nm)
{
ans[step*2]='\0';
flag=true;
return;
}
int tx,ty;
for(int i=0;i<8;i++)
{
tx=x+dir[i][0];
ty=y+dir[i][1];
if(tx<0||ty<0||tx>=n||ty>=m)
continue;
if(map1[tx][ty])
{
ans[step*2]=ty+65;
ans[step*2+1]=tx+49;
map1[tx][ty]=false;
record[tx][ty].x=x;
record[tx][ty].y=y;
DFS(tx,ty,step+1);
if(flag)
return;
map1[tx][ty]=true;
}
}
}
int main()
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
scanf("%d%d",&n,&m);
memset(map1,true,sizeof(map1));
nm=n*m;
flag=false;
map1[0][0]=false;
printf("Scenario #%d:\n",i);
DFS(0,0,1);
if(flag)
printf("%s\n\n",ans);
else
printf("impossible\n\n");
}
return 0;
}
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