标签：visit represent first struct bfs start make decide possible
POJ 2488 A Knight‘s Journey
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
题目大意：bfs求最短路问题，在棋盘上骑士要走遍棋盘上所有的位置，如果可以走完输出按字典序！！！排序的路径，否则输出impossible，这里特别注意国际象棋中骑士是按日字跳跃的，所以有8个方向可以行走，这里要输出按字典序排序的路径，那么可以直接找出所有可以到达的路径再进行排序，但是这样会超时，因为当棋盘很大的时候，走法很多。那么其实有个更巧妙的办法，那就是开始的时候把跳跃的8个方向的顺序按字典序排序，那么第一次成功时的路径就是字典序最小的路径。
int dx[] = {-1, 1, -2, 2, -2, 2, -1, 1};//按字典顺序定义好行走方向
int dy[] = {-2, -2, -1, -1, 1, 1, 2, 2};
代码：
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N = 30;
typedef struct {
int x;
int y;
} P;
P que[1000];
int cur = 0;
int n, m;
int vis[N][N];
int dx[] = {-1, 1, -2, 2, -2, 2, -1, 1};//按字典顺序定义好行走方向
int dy[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dfs(int x, int y, int num) {
int flag = 0;
if(num == n*m) {
printf("A1");
for(int i = 0; i < cur; i++) {
printf("%c%d", ‘A‘+que[i].y, que[i].x + 1);
}
printf("\n\n");
return 1;
}
for(int i = 0; i < 8; i++) {
int nx = x + dx[i], ny = y + dy[i];
if(nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny]) {
vis[nx][ny] = 1;
que[cur].x = nx;
que[cur].y = ny;
cur++;
flag = dfs(nx, ny, num+1);
vis[nx][ny] = 0;
cur--;
if(flag) return flag;
}
}
return flag;
}
int main() {
int t;
scanf("%d", &t);
for(int i = 1; i <= t; i++) {
scanf("%d%d", &n, &m);
vis[0][0] = 1;
printf("Scenario #%d:\n", i);
if(!dfs(0, 0, 1)) {
printf("impossible\n\n");
}
}
return 0;
}
POJ 2488 A Knight's Journey
标签：visit represent first struct bfs start make decide possible
原文地址：https://www.cnblogs.com/kindleheart/p/9296931.html
