dfs 注意字典序输出
|
A Knight's Journey
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 Sample Output Source
TUD Programming Contest 2005, Darmstadt, Germany
|
[Submit] [Go Back] [Status] [Discuss]
/*=============================================================================
# FileName: 2488.cpp
# Desc: poj 2488
# Author: zhuting
# Email: cnjs.zhuting@gmail.com
# HomePage: my.oschina.net/locusxt
# Version: 0.0.1
# CreatTime: 2013-12-19 22:46:13
# LastChange: 2013-12-19 22:46:13
# History: dfs
=============================================================================*/
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#define maxn 66
int x_size = 0, y_size = 0;
int chess_num = 0;/*棋盘的格子数*/
int rout[maxn] = {0};/*路径*/
bool color[10][10] = {0};/*记录是否走过*/
int col_sum = 0;/*走过的点的总数*/
const int del_x[8] = {-2, -2, -1, -1, 1, 1, 2, 2};/*路径要字母序*/
const int del_y[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
bool dfs(int x, int y)
{
if (x < 0 || x >= x_size || y < 0 || y >= y_size)/*越界*/
return 0;
if (color[x][y])/*已经走过*/
return 0;
color[x][y] = 1;
++col_sum;
rout[col_sum - 1] = x * y_size + y;
if (col_sum == chess_num)/*已经遍历全部*/
return 1;
for (int i = 0; i < 8; ++i)
{
bool is_pos = dfs(x + del_x[i], y + del_y[i]);
if (is_pos) return 1;
}
--col_sum;
color[x][y] = 0;
return 0;
}
void init()
{
memset(color, 0, sizeof(color));
memset(rout, 0, sizeof(rout));
chess_num = x_size * y_size;
col_sum = 0;
}
int main()
{
int test_num = 0;
scanf("%d", &test_num);
for (int i = 0; i < test_num; ++i)
{
printf("Scenario #%d:\n", i + 1);
scanf("%d %d", &y_size, &x_size);/*写完发现搞反了*/
init();
if (dfs(0, 0))
{
for (int j = 0 ; j < chess_num; ++j)
{
printf("%c%d", rout[j] / y_size + 'A', rout[j] % y_size + 1);
}
printf("\n\n");
continue;
}
printf("impossible\n\n");
}
return 0;
}
来源：oschina
链接：https://my.oschina.net/u/1053833/blog/186107