Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 29241 | Accepted: 10027 |
Description
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
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};
它仅仅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; }