POJ2488 A Knight's Journey(AC2)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 36713 Accepted: 12469

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

测试数据可见http://poj.org/showmessage?message_id=162294好的分析路径：http://blog.csdn.net/lyy289065406/article/details/6647666花费时间：一天，还是有参考百度，自己要增加独立思考的能力 

注意国际象棋中行为数字，列为字母

 
 
//第二次做的
#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>//POJ2488 A Knight's Journey#if 1//这个顺序也是要严格按照字典序的顺序来的，要不然得到的结果就是不对的//重新做这道题主体代码都是对的，主要就是这个顺序没弄对，还有这个//注意马到1的位置是(-1,-2)不是(-2,-1)这个自己是严重犯了错误
//同样输出少了个\n
#define MAXINT 30int dir[8][2] = { { -1, -2 }, { 1, -2 }, { -2, -1 }, { 2, -1 }, { -2, 1 }, { 2, 1 }, { -1, 2 }, { 1, 2 } };char path[MAXINT][2]; //有2个长度A1就是2个长度int visit[MAXINT][MAXINT];int p = 0;int q = 0;void init(){int i = 0;int j = 0;for (i = 0; i < MAXINT;i++){for (j = 0; j < MAXINT; j++){visit[i][j] = 0;}path[MAXINT][0] = '\0';path[MAXINT][1] = '\0';}return;}void printfpath(){int i = 0;for (i = 1; i <= (p*q); i++){printf("%c%c", path[i][0], path[i][1]);}printf("\n");return;}//x负责数字，y负责字母，第1列ychar GetnameZimu(int y){return ('A' + y - 1);}char GetnameShuzi(int x){return ('0' + x);}//p是列,p负责数字，q是行，q负责字母int dfs(int x, int y,int num){int i = 0;if ((x<1)||(x>p)||(y<1)||(y>q)) return 0;if ((p*q) == num) return 1;for (i = 0; i < 8;i++){int x1 = x + dir[i][0];int y1 = y + dir[i][1];if ((x1<1) || (x1>p) || (y1<1) || (y1>q)) continue;if (1 == visit[x1][y1]) continue;visit[x1][y1] = 1;path[num + 1][0] = GetnameZimu(y1);path[num + 1][1] = GetnameShuzi(x1);if (1 == dfs(x1,y1,num+1)){return 1;}else{//回溯visit[x1][y1] = 0;path[num + 1][0] = '\0';path[num + 1][1] = '\0';}}return 0;}int main(){int i = 0;int T = 0;freopen("input.txt","r",stdin);scanf("%d",&T);for (i = 0; i < T;i++){init();scanf("%d %d", &p, &q);//可以从任何一个位置出发，那就从第一个位置出发，这个是(1,1)开始，这个是字典序中最小的printf("Scenario #%d:\n", i + 1);visit[1][1] = 1;path[1][0] = 'A';path[1][1] = '1';if (1 == dfs(1, 1,1)){printfpath();}else{printf("impossible\n");}printf("\n");}return 0;}#endif

 

//code pojAC

#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>/*31 12 34 3从题意中看4 3, 4是列，4是行*//*提交记录1：第一次提交wa, 再看了下题目加百度，发现输入p,q不是那么简单理解的,自己理解成p是列，q是行，不对的，应该是p是行，q是列2：第二次提交提示格式错误，原来输出要隔2行，好吧。。。*/#define MAXINT 100typedef struct info{ char zimu; int index;}infos;int p        = 0;int q       = 0;int visitsqure = 1;int sum = 0;int flag = 0;int   map[MAXINT][MAXINT];char  zimu[MAXINT][MAXINT];int   visit[MAXINT][MAXINT];infos que[MAXINT*MAXINT];//因为输出要求按照字典序，因此这个方向也是要按照字典序的方向,先拍字母，然后排数字，这个方向已定要对int dir[8][2] = { { -1, -2 }, {1,-2}, {-2,-1}, {2,-1}, {-2,1}, {2,1}, {-1,2}, {1,2} };int dfs(int x, int y, int step){ int i = 0; if (x <= 0 || x > p || y<0 || y>q) return 0; visit[x][y] = 1;  que[step].zimu = zimu[x][y]; que[step].index = map[x][y]; if (1 == flag) return 1; if (step == sum) {  flag = 1;  return 1; } for (i = 0; i < 8;i++) {  int x1 = x + dir[i][0];  int y1 = y + dir[i][1];  if (x1 <= 0 || x1 > p || y1<=0 || y1>q) continue;  if (visit[x1][y1]) continue;  visit[x1][y1] = 1;  if (1 == dfs(x1, y1, step+1))  {   return 1;  }  visit[x1][y1] = 0; } return 0;}void init(){ int k = 0; int j = 0; int tmp = 1;  for (j = 1; j <= p; j++) {  for (k = 1; k <= q; k++)  {   map[j][k]   = 0;   zimu[j][k]  = '\0';   visit[j][k] = 0;  }  que[tmp].zimu    = '\0';  que[tmp++].index = 0; } visitsqure = 1; sum = 0;    flag = 0; return;}//从第一点出发int main(){ int T = 0; int i = 0; int j = 0; int k = 0; int zimuindex = 65; //小a 97，大A65 freopen("input.txt", "r", stdin); scanf("%d",&T); for (i = 0; i < T;i++) {  scanf("%d %d", &p, &q);  init();  sum = p*q;  zimuindex = 65;  for (j = 1; j <= q;j++)  {   for (k = 1; k <= p; k++)   {    map[k][j]  = k;    zimu[k][j] = zimuindex; //输入也要注意，之前自己就弄错了,输入好像还是弄错了，p是行，q是列，自己以为输入是列，行，不对，是行，列   }   zimuindex += 1;  }  //要求是按照字典序的顺序输出的  if (1 == dfs(1, 1, 1))//从1行1列经过  {   printf("Scenario #%d:\n", i + 1);   for (j = 1; j <= sum;j++)   {    printf("%c%d", que[j].zimu, que[j].index);   }   printf("\n\n");  }  else  {   printf("Scenario #%d:\n",i+1);   printf("impossible\n\n");  } } return 0;}