A Knight's Journey（dfs+最小字典序）
A - A Knight's Journey
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Appoint description: System Crawler (2015-10-14)
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.
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.
If no such path exist, you should output impossible on a single line.
Sample Input
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
这题的搜索不难，难在字典序。而看了看题解后发现是有规律的。只要每次dfs要从每列这样遍历，而且方向数组要从左到右，再从上到下的次序就能保证是最小的字典序了。
AC代码：
#include
#include
#include
#include
#include
#include
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
vector< pair > k;
pair v;
int n,m,flag,cnt;
int fx[][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1}
,{-2,1},{2,1},{-1,2},{1,2}};
int map[55][55];
bool jugde(int x,int y)
{
if(x>=1&&x<=n&&y>=1&&y<=m&&!map[x][y]){
return true;
}
return false;
}
void dfs(int x,int y,int c)
{
int tx,ty;
if(c==n*m){flag=0;}
if(!flag)return;
for(int i=0;i<8;++i){
v.first=x, tx = fx[i][0] + x;v.first=tx;
v.second=y, ty = fx[i][1] + y;v.second=ty;
if(jugde(tx,ty)){
k.push_back(v);map[tx][ty]=1;
dfs(tx,ty,c+1);
if(!flag)return;
k.pop_back();
map[tx][ty]=0;
}
}
}
int main()
{
/*freopen("input.txt","r",stdin);*/
int N,i,j,c=0;
scanf("%d",&N);
while(N--)
{
flag = 1;cnt=0;
scanf("%d%d",&n,&m);
printf("Scenario #%d:\n",++c);
for(i=1;i<=m&&flag;++i){
for(j=1;j<=n&&flag;++j){
map[j][i] = 1;//这里只是换了m，n位置而未换i，j所以wa了
v.first = j,v.second = i;
k.push_back(v);
dfs(j,i,1);
CRL(map);
if(flag)
k.clear();
}
}
if(!flag){
for(i=0;i