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POJ2488:A Knight's Journey(dfs)
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.
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.
If no such path exist, you should output impossible on a single line.
Sample Input
31 12 34 3
Sample Output
Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4
题目大意:
给出国际象棋的的一个棋盘,里面有一个马,给出所在棋盘的长和宽,问从任意一点开始,能否不重复的走遍棋盘上的每一点,若能就输出一个字符串,其中字母代表行,数字代表列(按字典序搜索),不能就输出impossible
解题思路:
很明显可以用DFS暴力搜索,但是要注意的是,骑士在搜索的时候按字典序的方向搜索。
题目解析:
一直没读懂题,做出来的答案一直和样例不同,后来看题解,说是要按字典序搜索,然后又是N遍WA,(只能说dfs自己学的很渣,递归一会就
递晕了),然后没敢深入思考,别人的代码,都是指dfs了一遍,即dfs(1,1,1),我不知道为什么,只好枚举所有结果,综合来说,算是一道
水题吧,只要知道按字典序搜索就好了。
#include <stdio.h>#include <iostream>#include <string.h>#include <stdlib.h>using namespace std;int jx[8]= {-2,-2,-1,-1,1,1,2,2};int jy[8]= {-1,1,-2,2,-2,2,-1,1};int m,n,v[28][28],flag;char f[28*28];int g[28*28];void pu(){ for(int i=1; i<=n*m; i++) { printf("%c%d",f[i],g[i]); } printf("\n");}void dfs(int x,int y,int ans){ int tx,ty; if(ans==n*m) { f[ans]=x+‘A‘-1; g[ans]=y; flag=1; return ; } for(int i=0; i<8; i++) { tx=x+jx[i]; ty=y+jy[i]; if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&v[tx][ty]==0) { v[tx][ty]=1; f[ans+1]=tx+‘A‘-1; g[ans+1]=ty; dfs(tx,ty,ans+1); if(flag==1) return ; v[tx][ty]=0; } } return ;}int main(){ int T; scanf("%d",&T); for(int z=1; z<=T; z++) { flag=0; scanf("%d%d",&m,&n); printf("Scenario #%d:\n",z); for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) { memset(v,0,sizeof(v)); v[i][j]=1; f[1]=i-1+‘A‘; g[1]=j; dfs(i,j,1); if(flag==1) { pu(); break; } } if(flag==1) break; } if(flag==0) printf("impossible\n"); printf("\n"); } return 0;}
POJ2488:A Knight's Journey(dfs)
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