深搜基础

最新推荐文章于 2025-08-14 20:57:38 发布

kidozh

最新推荐文章于 2025-08-14 20:57:38 发布

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分类专栏： DFS 文章标签：算法

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本文探讨如何通过深度搜索算法解决骑士在不同大小棋盘上进行一次且仅一次访问所有方格的问题，并提供了实现代码及示例输入输出。

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Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

—————————————————————————————————————————————————————————————————————————————

分界线

这个题很明显是深搜的问题第一次wa是因为一些细节原因（复制粘贴惹的祸）第二次是字母表的顺序没调对，最后一次得怪可恶的空行！！！！！！！！！！！

下面就是代码了注意字母表的规律的时候一定要注意+2 +1的细节之差.....（说多了都是泪）

#define _CRT_SECURE_NO_WARNINGS//vs防报错  scanf不安全....
#include<cstdio>
#include<cstring>

int vis[100][100];
int a, b, i, j, flag;
void trav(int x, int y, int n)
{
	int o;
	if (flag == 1)return;//一旦找到就退出
	if (vis[x][y] == 0 && x >= 1 && x <= a&&y >= 1 && y <= b){//判定是否在局内
		if (n >= a*b&&flag != 1){//找到了就出结果
			vis[x][y] = n;
			flag = 1;
			for (o = 1; o <= a*b; o++){
				for (i = 1; i <= a; i++){
					for (j = 1; j <= b; j++){
						if (vis[i][j] == o)printf("%c%d", i + 'A' - 1, j);//vis记录数据 这样可能导致计算的时候效率低下 可以采用struct来弥补这个问题
					}
				}
			}
			printf("\n");
		}
		vis[x][y] = n;//标记为已经寻的
		n++;
		trav(x - 2, y - 1, n);
		trav(x - 2, y + 1, n);
		trav(x - 1, y - 2, n);
		trav(x - 1, y + 2, n);
		trav(x + 1, y - 2, n);
		trav(x + 1, y + 2, n);
		trav(x + 2, y - 1, n);
		trav(x + 2, y + 1, n);//8方向遍历
		vis[x][y] = 0;//如果什么都不成功 返回到上一步
	}
	return;
}
int main()
{
	int i, n;
	//freopen("test.txt", "r", stdin);
	scanf("%d", &n);
	for (i = 0; i<n; i++){
		scanf("%d%d", &b, &a);
		flag = 0;
		printf("Scenario #%d:\n", i + 1);
		memset(vis, 0, sizeof(vis));
		trav(1, 1, 1);
		if (flag == 0)printf("impossible\n");
		if(i!=n-1)printf("\n");//唯有泪千行！！！！！！！！一定！！！！！！
	}
	return 0;
}

深搜 基础

深搜基础