Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
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
横纵坐标。。。。。。怎么分辨
字典序要求搜索时
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
If no such path exist, you should output impossible on a single line.
3 1 1 2 3 4 3Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4代码:
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int maxn=10010;
int v[35][35],v1[35][35];
char vis[905][2];
int f[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int l=1,n,l1,l2,flag,t=1;
int Judge(int x1,int y1)
{
if(x1<0||x1>=l1||y1<0||y1>=l2)return 0;
if(v[x1][y1]>0)return 0;
return 1;
}
void dfs(int x,int y)
{
vis[l][0]=x+'1';
vis[l][1]=y+'A';
if(l==l1*l2){
flag=1;
for(int i=1;i<=l1*l2;i++){
printf("%c%c",vis[i][1],vis[i][0]);
}
printf("\n");
return;
}
for(int k=0;k<8&&(flag==0);k++){
int x1,y1;
x1=x+f[k][0];
y1=y+f[k][1];
if(Judge(x1,y1)){
v[x1][y1]=1;
l++;
dfs(x1,y1);
l--;
v[x1][y1]=0;
}
}
return;
}
int main(){
scanf("%d",&n);
while(n--){
scanf("%d %d",&l1,&l2);
printf("Scenario #%d:\n",t++);
l=1; flag=0;
v[0][0]=1;
dfs(0,0);
memset(v,0,sizeof(v));
memset(v1,0,sizeof(v1));
memset(vis,0,sizeof(vis));
if(!flag) printf("impossible\n");
if(n) printf("\n");
}
return 0;
}
在搜索是直接从(0,0)开始就行,在某一点存在不能搜索的点,在任何地方都会有不能搜索到的点。
横纵坐标。。。。。。怎么分辨
字典序要求搜索时
f[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
一定是向坐标最小的地方走