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poj 2488 -- A Knight's Journey

2024-07-15 07:17:28   214人阅读

A Knight‘s Journey
 
 
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 30388 Accepted: 10404

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题目大意： 让骑士以字典序遍历整个图，每个点只能到一次。

思路：数据不大，直接搜索即可。注意每个测试数据后面都有一个空行。

 1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   AKnightsJourney.cpp 4  *       Creat time :   2014-07-31 16:22 5  *      Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 3015 using namespace std;16 struct Node17 {18     int x,y;19 }steps[M*M];20 bool vis[M][M];21 int p,q;22 int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};23 int DFS(int x,int y,int cnt)24 {25     if(vis[x][y]){26         return -1;27     }28     vis[x][y] = true;29     steps[cnt].x = x; steps[cnt].y = y;30     if(cnt == p*q-1) return 1;31     for(int i = 0; i < 8; i++){32         int xx = x+dir[i][0];33         int yy = y+dir[i][1];34         if(xx >= 1 && xx <= p && yy >= 1 && yy <= q){35             int ans = DFS(xx,yy,cnt+1);36             if(ans == 1){37                 return true;38             }39             else if(ans == 0){40                 vis[xx][yy] = false;41             }42         }43     }44     return 0;45 }46 int main(int argc,char *argv[])47 {48     int n,t = 1;49     scanf("%d",&n);50     while(n--){51         clr(steps,0);52         scanf("%d%d",&p,&q);53         int cnt = 0;54         printf("Scenario #%d:\n",t++);55         int flag = 0;56         for(int i = 1; i <= q; i++){57             for(int j = 1; j <= p; j++){58                 clr(vis,0);59                 if(DFS(j,i,cnt) == 1){  //因为i与j位置反了，贡献了几次wr60                     flag = 1;61                     break;62                 }63             }64             if(flag) break;65         }66         if(!flag){67             printf("impossible");68         }69         else{70             for(int i = 0; i < q*p; i++){71                 printf("%c%d",steps[i].y-1+‘A‘,steps[i].x);72             }73         }74         printf("\n\n");75     }76     return 0;77 }
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