A Knight's Journey
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 29241 |
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Accepted: 10027 |
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
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
#include
#include
#include
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; }