A Knight's Journey
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 31147 |
|
Accepted: 10655 |
Description
Background
The knight is getting bZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"pst">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
3
1 1
2 3
4 3
Sample Output
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
Source
TUD Programming Contest 2005, Darmstadt, Germany 给出m行n列的棋盘，问骑士能不能走完所有的点，每个点只能走一次，要求字典序最小输出。 如果能走完，那么一定可以从A1开始走，字典序最小 只要dfsA1开始，看能不能走完，注意搜索的顺序。
#include
#include
#include
using namespace std; int t , n , m , sum ; int mm[30][30] ; int pre[1000] ; int a[8][2] = { {-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1} }; int dfs(int x,int y,int temp) { if( temp == sum ) return 1 ; int flag = 0 , xx , yy , i ; for(i = 0 ; i < 8 ; i++) { xx = x + a[i][0] ; yy = y + a[i][1] ; if( xx >= 0 && xx < n && yy >= 0 && yy < m && !mm[xx][yy] ) { mm[xx][yy] = 1 ; pre[x*m+y] = xx*m+yy ; flag = dfs(xx,yy,temp+1); if( flag ) return flag ; mm[xx][yy] = 0 ; } } return flag ; } int main() { int i , j , k , tt ; scanf("%d", &t); for(tt = 1 ; tt <= t ; tt++) { memset(mm,0,sizeof(mm)); memset(pre,-1,sizeof(pre)); scanf("%d %d", &m, &n); sum = n*m ; mm[0][0] = 1 ; k = dfs(0,0,1); printf("Scenario #%d:\n", tt); if(k == 0) printf("impossible"); else { for(i = 0 , k = 0 ; i < sum ; i++) { printf("%c%c", k/m+'A', k%m+'1'); k = pre[k] ; } } printf("\n\n"); } return 0; }