骑士游历问题
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
题意分析:
基础dfs暴力,类似于马走日问题,但是要注意几个问题:
1.判断不能全部遍历完的情况 输出impossible
2.每一次输出要输出两个“\n”
3.输出第一次以字典序最小排序的遍历情况
代码:
#include<stdio.h>
#include<stdbool.h>
#include<string.h>
//通过控制遍历方向(前进一格优先,后退两个优先)使字典序最小排序
int solution_x[8]= {-1,1,-2,2,-2,2,-1,1};
int solution_y[8]= {-2,-2,-1,-1,1,1,2,2};
int a[1000];
char b[1000];
int T,n,m,flag;
int chess[10][10]= {0}; //模拟棋盘
int count;
void SetChess(int step,int x1,int y1)
{
int i;
if(step==n*m)
{
if(count==0) //count记录第几个解 若是第一个 则输出
{
printf("A1");
for(i=1; i<step; i++)
printf("%c%d",b[i],a[i]);
printf("\n\n");
flag=1;
count++;
}
return;
}
for (i=0; i<8; i++)
{
int x=x1+solution_x[i],y=y1+solution_y[i];
if (chess[x][y]==0&& x>0&&y>0 && x<=n&&y<=m)
{
chess[x][y]=1;
a[step]=x,b[step]='A'+y-1;
SetChess(step+1,x,y);
chess[x][y]=0;
}
}
}
int main()
{
scanf("%d",&T);
int number=1;
while (T--)
{
count=0;
flag=0;
scanf("%d %d",&n,&m);
printf("Scenario #%d:\n",number++);
memset(chess,0,sizeof(chess));
chess[1][1]=1;
SetChess(1,1,1);
if(flag==0)
printf("impossible\n\n");
}
}