UVA 439 (骑士移动 第一道DFS 13.07.17)
2013-07-18 12:29
381 查看
Knight Moves |
difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplishedthis, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route froma tob.
Input Specification
The input file will contain one or more test cases. Each test case consists of one line containing twosquares separated by one space. A square is a string consisting of a letter (a-h) representing thecolumn and a digit (1-8) representingthe row on the chessboard.
Output Specification
For each test case, print one line saying "To get from xx toyy takes n knight moves.".
Sample Input
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves. 题意不难理解, 从棋盘上的一个位置倒另一个位置 需要的最少步数~ AC代码如下:
#include<stdio.h> #define N 250 int dir[8][2] = {{-1, 2}, {-2, 1}, {-2, -1}, {-1, -2}, {1, -2}, {2, -1}, {2, 1}, {1 ,2}}; char sch, ech; int s, e; int used ; void DFS(char x, int y,int cnt) { if(x < 'a' || x > 'h' || y < 1 || y > 8 ||used[x][y] <= cnt) return ; used[x][y] = cnt; for(int i = 0; i < 8; i++) { int tx = x + dir[i][0]; int ty = y + dir[i][1]; DFS(tx, ty, cnt+1); } } int main() { while(scanf("%c%d", &sch, &s) != EOF) { getchar(); scanf("%c%d", &ech, &e); getchar(); for(int i = 0; i < N; i++) for(int j = 0; j < N ; j++) used[i][j] = 65; DFS(sch, s, 0); printf("To get from %c%d to %c%d takes %d knight moves.\n", sch, s, ech, e, used[ech][e]); } }
相关文章推荐
- uva 439 Knight Moves 骑士移动
- 习题6-4 骑士的移动 UVa 439
- uva 439骑士的移动(图的最短路 bfs)
- 习题 6-4 UVA 439 Knight Moves 骑士的移动
- 习题6-4 骑士的移动(Knight Moves, UVa 439)
- UVA 439 BFS 骑士的移动
- uva 439 Knight Moves(马移动) —— DFS + 剪枝
- UVa 439骑士的移动(BFS)
- Uva 439 骑士的移动
- uva 439 Knight Moves(骑士的移动)BFS
- 骑士的移动搜索 UVA439
- uva-439 骑士的移动
- uva 439 Knight Moves 骑士移动
- 骑士的移动(Knight Moves, UVa 439)
- 习题6-4 骑士的移动 UVa439
- 骑士的移动(Knight Moves,UVa 439)
- UVA 439 Knight Moves --DFS or BFS
- UVA 439 Knight Moves 走象棋 (DFS or BFS)
- Uva 439 - Knight Moves(骑士游历)
- UVA 439 骑士游历