标签：square Scenario int Knight dfs Journey knight path line
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.
无聊了 想旅行
Whenever a knight moves, it is two squares in one direction and one square perpendicular to this.
走法：一个方向的2个方格，垂直于此的一个正方形。（其实就是走日！）
The world of a knight is the chessboard he is living on.
世界就是他生活的棋盘
Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board,
我们国王生活在棋盘上,有一个 比常规8 * 8棋盘小的区域。
but it is still rectangular.
依然是矩形
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.
遍历每一个地方，可以在任何地方的棋盘 开始/结束
Input
The input begins with a positive integer n in the first line.
The following lines contain n test cases.
n个测试用例
Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26.
每个用例是 p q 并且 p、q >= 1 q<=26
This represents a p * q chessboard,
这代表p·* q棋盘
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
注意看好，这个顺序 非常重要，否则wa！
x代表abcd y轴代表 1234
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
typedef struct Path {
int y;
int x;
} Path;
Path path[1001];
bool isFound = false;
int p,q;
bool mark[51][51];
int dir[8][2] = {
{-1,-2},
{1,-2},
{-2,-1},
{2,-1},
{-2,1},
{2,1},
{-1,2},
{1,2},
};
// p * q
bool isInMap(int ny,int nx) {
if (nx >= 1 && nx <= q && ny >= 1 && ny <= p) {
return 1;
}
return 0;
}
void dfs(int y,int x,int step) {
// cout << "current:" << y <<" "<< x << endl;
if (step == p * q) {
// cout << "out!!! :" << step << endl;
for (int i = 1; i <= step; i ++) {
cout << char(path[i].x + 'A' - 1) << path[i].y;
}
cout << endl;
isFound = 1;
return ;
}
int ny;
int nx;
for (int i = 0; i <= 7; i++) {
ny = y + dir[i][0];
nx = x + dir[i][1];
if (isFound == 0) {
if (isInMap(ny,nx) == 1 && mark[ny][nx] == 0) {
path[step+1].y = ny;
path[step+1].x = nx;
mark[ny][nx] = 1;
// dfs
dfs(ny,nx,step+1);
// 回溯
mark[ny][nx] = 0;
}
}
}
}
int main() {
int caseNum;
cin >> caseNum;
for (int caseNo = 1; caseNo <= caseNum; caseNo++) {
cin >> p >> q;
isFound = false;
for (int i = 1; i <= p ; i++) {
for (int j = 1; j <= q; j++) {
mark[i][j] = 0;
}
}
path[1].y = 1;
path[1].x = 1;
mark[1][1] = 1;
cout << "Scenario #" << caseNo << ":" <<endl;
dfs(1,1,1);
if (!isFound) {
cout << "impossible" << endl << endl;
} else {
cout << endl;
}
}
}
标签：square,Scenario,int,Knight,dfs,Journey,knight,path,line 来源： https://blog.csdn.net/bijingrui/article/details/104782732
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。