A Knight's Journey
2013-11-14 19:23
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A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 26808 | Accepted: 9140 |
Background
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
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
题目大意:就是让骑士走遍棋盘, 按日字形走, 找出可遍历棋盘的路径, 并按照字典序输出
思路:主要都是按字典序输出, 这与方向的优先级别有关, 还有就是每组数据结束后有一空行, 不要被坑了。。。
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> using namespace std; #define MAX 10 bool vis[MAX][MAX], flag; //标记已访问,是否查找到结果 int m, n, cnt = 1, dir[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}, {2, -1}, {2, 1}}; //方向, 注意如果按字典序输出的话必须x从小到大,y从小到大 typedef struct Elem //用来记录路径的结构体 { int x; int y; }elem; elem ans[100]; //用来记录路径的结构体 void dfs(int count, elem pos) //用深搜搜出路径 { if(flag) { return; } elem temp; if(count == m * n) //找到路径, 打印结果 { printf("Scenario #%d:\n", cnt++); for(int i = 0; i < m*n; i++) { printf("%c%d", ans[i].x+65, ans[i].y+1); } printf("\n\n"); //注意此处超坑要加空行 flag = 1; return; } for(int j = 0; j < 8; j++) { temp.x = pos.x + dir[j][0]; temp.y = pos.y + dir[j][1]; if(temp.x >= 0 && temp.x < n && temp.y >= 0 && temp.y < m && !vis[temp.x][temp.y]) //不多说老套深搜 { ans[count] = temp; vis[temp.x][temp.y] = 1; dfs(count+1, temp); vis[temp.x][temp.y] = 0; } } } int main() { int k; elem sp = {0, 0}; //起点 scanf("%d", &k); while(k--) { scanf("%d%d", &m, &n); memset(vis, 0, sizeof(vis)); flag = 0; ans[0] = sp; vis[0][0] = 1; dfs(1, sp); if(!flag) //未找到路径 { printf("Scenario #%d:\nimpossible\n\n", cnt++); } } return 0; }
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