poj 2488 A Knight's Journey(dfs+字典序路径输出)
2014-07-16 15:44
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题目链接:http://poj.org/problem?id=2488
Description
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
Sample Output
大致题意:
给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径。
代码如下:
题目链接:http://poj.org/problem?id=2488
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Description
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
大致题意:
给出一个国际棋盘的大小,判断马能否不重复的走过所有格,并记录下其中按字典序排列的第一种路径。
代码如下:
#include <cstdio> #include <cstring> #define M 26 struct node { int x, y; }w[M*M]; bool vis[M][M]; int p, q; int flag = 0; int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}}; //按此顺序搜索出来的结果就是字典序 bool judge(int x, int y) { if(x>=0&&x<q&&y>=0&&y<p&&!vis[x][y]) return true; return false; } void dfs(int x, int y, int step) { w[step].x = x,w[step].y = y; vis[x][y] = true; if(step == p*q-1) { flag = 1; return ; } for(int i = 0; i < 8; i++) { int dx = w[step].x+dir[i][0]; int dy = w[step].y+dir[i][1]; if(judge(dx,dy)) { vis[dx][dy] = true; dfs(dx,dy,step+1); if(flag)//一但找到就退出搜索 return; vis[dx][dy] = false; } } return ; } void print() { for(int i = 0; i < p*q; i++)//列为字母,行为数字 { printf("%c%d",w[i].x+'A',w[i].y+1); } printf("\n\n"); } int main() { int t, i, j, cas = 0; scanf("%d",&t); while(t--) { memset(vis,false,sizeof(vis)); flag = 0; scanf("%d%d",&p,&q); for(i = 0; i < q; i++)//列 { for(j = 0; j < p; j++)//行 { dfs(i,j,0); if(flag) break; } if(flag) break; } printf("Scenario #%d:\n",++cas); if(flag) print(); else printf("impossible\n\n"); } return 0; }
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