poj_2488 A Knight's Journey（dfs）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42365 Accepted: 14404

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

注意要字典序输出，所以dfs的顺序可以规定下。当然因为数据量小，每次找到新路后都判断字典序一下也是可以过的。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 1010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Point{    int x, y;}path[maxn];int n, m;int maxc;bool flag;int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};bool vis[30][30];void dfs(int x, int y, int cot){    if(flag) return;        vis[x][y] = true;    path[cot].x = x, path[cot].y = y;    if(cot == maxc)    {        flag = 1;        return ;    }    for(int i = 0; i < 8; i++)    {        int xx = x + dx[i], yy = y + dy[i];        if(xx < 1 || xx > n || yy < 1 || yy > m) continue;        if(!vis[xx][yy])        {            dfs(xx, yy, cot+1);            vis[xx][yy] = false;        }    }}int main(){    int casen = 0;    int T;    scanf("%d", &T);    while(T--)    {        flag = 0;        casen++;        scanf("%d%d", &m, &n);        maxc = n * m;        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= m; j++)            {                memset(vis, false, sizeof(vis));                dfs(i, j, 1);                if(flag) break;            }            if(flag) break;        }        printf("Scenario #%d:\n", casen);        if(!flag) printf("impossible\n");        else        {            for(int i = 1; i <= maxc; i++) printf("%c%d", path[i].x + 'A' - 1, path[i].y);            printf("\n");        }        printf("\n");    }    return 0;}