A Knight's Journey (poj 2488 DFS)
Language:
A Knight's Journey
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 3 1 1 2 3 4 3 Sample Output Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4 Source
TUD Programming Contest 2005, Darmstadt, Germany
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题意:给你一个p*q的国际象棋棋盘,问马从任意一点出发能否每个点只经过一次把所有的点都遍历完。其实只需要从最左上角开始走就行了,因为如果每个点都能走到起点在哪就无所谓了。另外要按字典序输出,则dir数组是一定的。最后注意国际象棋行为数字列为字母,开始不小心搞反了一直不对。。。。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 2005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; struct Node { int x,y; }path[30]; int visit[30][30]; int p,q,step,flag; int dir[8][2]={-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2}; bool ISok(int x,int y) { if (!visit[x][y]&&x>=0&&x<p&&y>=0&&y<q) return true; return false; } void dfs(int x,int y) { if (step==p*q) { if (!flag) { for (int i=0;i<p*q;i++) printf("%c%d",path[i].y+'A',path[i].x+1); printf("\n"); flag=1; } return ; } for (int i=0;i<8;i++) { int dx=x+dir[i][0]; int dy=y+dir[i][1]; if (ISok(dx,dy)) { visit[dx][dy]=1; path[step].x=dx; path[step].y=dy; step++; dfs(dx,dy); visit[dx][dy]=0; step--; } } } int main() { int Q,cas=1; scanf("%d",&Q); while (Q--) { scanf("%d%d",&p,&q); memset(visit,0,sizeof(visit)); printf("Scenario #%d:\n",cas++); path[0].x=0; path[0].y=0; visit[0][0]=1; step=1; flag=0; dfs(0,0); if (!flag) printf("impossible\n"); printf("\n"); } return 0; } /* 3 1 1 2 3 4 3 */