Day2-F-A Knight's Journey POJ-2488
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
Output
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 简述:给你一副p*q的棋盘,求出每个点恰好只走一次的路径,若有多个答案输出字典序最小的。
分析:求最深路径,只经过一次,DFS+回溯,注意到如果其能满足题意,那么必然会经过A1,从A1开始字典序最小,所以从A1开始DFS搜索即可,注意保证dx与dy也是字典序排序,代码如下:
const int maxm = ;
//注意字典序大小排序
const int dx[] = {-, , -, , -, , -, };
const int dy[] = {-, -, -, -, , , , }; int vis[maxm][maxm], Next[maxm][maxm], r, c, n, kase = ; bool inside(int x,int y) {
return x > && x <= r && y > && y <= c;
} void print(int x,int y,int t) {
if(t) {
printf("%c%d", y - + 'A', x);
print(Next[x][y] / , Next[x][y] % , t - );
}
} bool dfs(int x,int y,int t) {
vis[x][y] = ;
if(t == r * c) {
return true;
}
for (int i = ; i < ; ++i) {
int nx = x + dx[i], ny = y + dy[i];
if(inside(nx,ny) && !vis[nx][ny]) {
Next[x][y] = nx * + ny;
if(dfs(nx, ny,t+)) {
return true;
}
}
}
vis[x][y] = ;
return false;
} int main() {
scanf("%d", &n);
while(n--) {
scanf("%d%d", &r, &c);
memset(vis, , sizeof(vis)), memset(Next, , sizeof(Next));
printf("Scenario #%d:\n", ++kase);
if(dfs(, , )) {
print(,,r*c);
} else {
printf("impossible");
}
printf("\n\n");
}
return ;
}
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