POJ 2488:A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29241 Accepted: 10027

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

这个道题超级坑，要注意字典数。

即

int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};

它只能按字典序顺序定义。。

还有就是每一轮输出后要加一个换行。。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int M = 100 + 5;int chess[M][M];int h[M];int l[M];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;int ok;void dfs(int x, int y, int ans){    h[ans]=x;    l[ans]=y;    if(ans==a*b)    {        ok=1;        return ;    }    for(int i=0; i<8; i++)    {        int tx = x + dx[i];        int ty = y + dy[i];        if(tx>=1 && tx<=b && ty>=1 && ty<=a && chess[tx][ty]==0 && ok==0)        {            chess[tx][ty] = 1;            dfs(tx, ty, ans+1);            chess[tx][ty] = 0;        }    }}int main(){    int n;    scanf("%d", &n);    for(int cas=1; cas<=n; cas++)    {        ok=0;        memset(chess, 0, sizeof(chess));        memset(h, 0, sizeof(h));        memset(l, 0, sizeof(l));        scanf("%d%d", &a, &b);        chess[1][1] = 1;        dfs(1, 1, 1);        printf("Scenario #%d:\n", cas);        if(ok==1)        {            for(int i=1; i<=a*b; i++)                printf("%c%d", h[i]+'A'-1, l[i]);            printf("\n");        }        else            printf("impossible\n");        printf("\n");    }    return 0;}