POJ 2488 A Knight's Journey (棋盘DFS)

A Knight's Journey

 

大意:

给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

 

 

#include <map>
#include <stack>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <limits.h>
#include <algorithm>
#define LL long long
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
#define eps 1e-9
#define INF 1 << 30
using namespace std;

bool vis[30][30], output;
int Num;
char path[120];
int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int a, b;

void Dfs(int depth, int x, int y)
{
    if(depth == Num)
    {
        for(int i = 0; i < 2*depth; i++)
            printf("%c", path[i]);
        printf("\n\n");
        output = true;
        return ;
    }
    for(int i = 0; i < 8 && output == false; i++)
    {
        int new_x = x+dx[i];
        int new_y = y+dy[i];
        if(new_x >= 1 && new_x <= b && new_y >= 1 && new_y <= a && vis[new_y][new_x] == false)
        {
            vis[new_y][new_x] = true;
            path[2*depth] = 'A'+new_x-1;
            path[2*depth+1] = '1'+new_y-1;
            Dfs(depth+1, new_x, new_y);
            vis[new_y][new_x] = false;
        }
    }
}

void run()
{
    int n;
    scanf("%d", &n);
    for(int p = 1; p <= n; p++)
    {
        scanf("%d%d", &a, &b);
        printf("Scenario #%d:\n", p);
        for(int i = 1; i <= a; i++)
            for(int j = 1; j <= b; j++)
                vis[i][j] = false;
        Num = a*b;
        output = false;
        vis[1][1] = true;
        path[0] = 'A';
        path[1] = '1';
        Dfs(1, 1, 1);
        if(output == false)
            printf("impossible\n\n");
    }
}

int main(void)
{
    run();

    return 0;
}