poj 2488 A Knight's Journey DFS
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A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 37746 Accepted: 12835
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
题意:就是说一个骑士能够走得有八个方向,让你找到一种方法可以走满整个棋盘。输出的:按照字典序输出
思路:DFS,深搜每一个点,知道能够完整走出来
WA点:1.字典序,确定了八个方向的值,必须是int xx[8] = {-2,-2,-1,-1,1,1,2,2}; int yy[8] = {-1,1,-2,2,-2,2,-1,1};这样的
2.国际棋盘横是字母,竖是数字,写程序一定要按照这样的顺序,否则答案对也是WA的
这样理解了就很容易AC了
AC代码:
#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;int n2,n1;int gra[30][30];int xx[8] = {-2,-2,-1,-1,1,1,2,2};int yy[8] = {-1,1,-2,2,-2,2,-1,1};int cas;int tag;char path[100];void dfs(int x,int y,int tol) { path[tol] = x+'A'-1; path[tol+1] = '0'+y; if((tol+2)/2 == n1*n2) { tag = 1; return; } for(int i = 0; i < 8; i++) { int g = x+xx[i]; int h = y+yy[i]; if(g >= 1 && g <= n1 && h >= 1 && h <= n2&& !gra[g][h] && !tag) { gra[g][h] = 1; dfs(g,h,tol+2); gra[g][h] = 0; } }}int main() { scanf("%d",&cas); for(int i = 1; i <= cas; i++) { scanf("%d%d",&n2,&n1); memset(gra,0,sizeof(gra)); memset(path,0,sizeof(path)); tag = 0; gra[1][1] = 1; dfs(1,1,0); printf("Scenario #%d:\n",i); if(tag) { printf("%s\n",path); } else { printf("impossible\n"); } if(i < cas) { printf("\n"); } } return 0;}
程序就是严密,之前写错八个方向,后来又搞错的横竖,虽然答案是对的,但就是WA哎!
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