A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 28697 | Accepted: 9822 |
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
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
#include <cstdio> #include <cstdlib> #include <stack> #include <cstring> using namespace std; const int MAX = 9; const int dirx[8]={-1,1,-2,2,-2,2,-1,1},diry[8]={-2,-2,-1,-1,1,1,2,2}; typedef struct Point{ int x,y; }point; int p,q,n; bool visit[MAX][MAX]; point pre[MAX][MAX]; bool mark; stack<int> stx,sty; void printPath(int x,int y){ stx.push(x); sty.push(y); int tx,ty; tx = pre[x][y].x; ty = pre[x][y].y; while(tx!=-1){ stx.push(tx); sty.push(ty); x = pre[tx][ty].x; y = pre[tx][ty].y; tx = x; ty = y; } while(!stx.empty()){ printf("%c%d",sty.top()-1+'A',stx.top()); stx.pop(); sty.pop(); } printf(" "); } void dfs(int x,int y,int len){ if(mark)return; if(len==p*q){ printPath(x,y); mark = true; return; } int i,tx,ty; for(i=0;i<8;++i){ tx = x+dirx[i]; ty = y+diry[i]; if(tx<1 || tx>p || ty<1 || ty>q)continue; if(visit[tx][ty])continue; pre[tx][ty].x = x; pre[tx][ty].y = y; visit[tx][ty] = true; dfs(tx,ty,len+1); visit[tx][ty] = false; } } int main() { //freopen("in.txt","r",stdin); //(Author : CSDN iaccepted) int i; scanf("%d",&n); for(i=1;i<=n;++i){ printf("Scenario #%d: ",i); scanf("%d %d",&p,&q); memset(visit,0,sizeof(visit)); mark = false; pre[1][1].x = -1; pre[1][1].y = -1; visit[1][1] = true; dfs(1,1,1); visit[1][1] = false; if(!mark){ printf("impossible "); } } return 0; }
题目意思:象棋中的马在一张棋盘上是否能不反复的走全然部格子。假设能走完输出走的路径(以字典序),假设没有一种走法能达到这种目标,则输出impossible。
思路就是DFS 搜下去,当走过的格子数达到格子总数时就打印路径。所以要用一个数组记录每一个定点的前驱节点。