2166: A Knight's Journey

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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

先发个图自嘲一下/(ㄒoㄒ)/~~真心羸弱

就先说刚开始，三个TLE，主要还是输入输出搞得，毕竟C++的输入输出的时间比较多，再加上暴力题，输出结果又多，╮(╯▽╰)╭

改完输入输出之后就开始WA，主要被坑的就是

1.对行列理解错了，把行和列弄反了/(ㄒoㄒ)/~~

2.就是字典序， lexicographically 英语不好还是坑了

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int p;
int q;
int visited[31][31];
bool flag;
char pathX[31];
int pathY[31];

/*字典序
 3    5
1      7                 
  
2      8
 4    6
*/
//1234
int dx[] = {-1,1,-2,2,-2,2,-1,1};
//ABCD
int dy[] = {-2,-2,-1,-1,1,1,2,2};


bool isSafe(int x,int y) {
    if(x >= 1 && x <= p && y >= 1 && y <= q && visited[x][y] == 0 && flag == false) {
        return true;
    }
    return false;
}

void dfs(int x,int y,int step) {
    if(flag == true) {
        return;
    }
    
    pathX[step] = x;
    pathY[step] = y + 'A' - 1;
    if(step == p * q) {
        flag = true;
        return;
    }
    
    for(int i = 0;i < 8;i++) {
        int nextX = x + dx[i];
        int nextY = y + dy[i];
        if(isSafe(nextX,nextY) == true) {
            visited[nextX][nextY] = 1;
            
            dfs(nextX,nextY,step+1);
            visited[nextX][nextY] = 0;
        }
    }
}

int main(void) {
    int n;
    int count = 1;
    
    scanf("%d",&n);
    
    while(n--) {
        flag = false;
        memset(visited,0,sizeof(visited));
        scanf("%d %d",&p,&q);
        printf("Scenario #%d:\n",count);
        visited[1][1] = 1;
        dfs(1,1,1);
        
        if(flag == false) {
            printf("%s\n","impossible");
        } else {
            for(int i = 1;i < p*q;i++) {
                printf("%c%d",pathY[i],pathX[i]);
            }
            printf("%c%d\n",pathY[p*q],pathX[p*q]);
        }

        printf("\n");

        count++;
    }
    
    return 0;
}