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?
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
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, …
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 Input31 12 34 3Sample OutputScenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4字典序，只要控制轉移方向的順序即可
#include<stdio.h>#include<string.h>#define MAX_N 30bool vis[MAX_N][MAX_N];int W,H,goal;int oy[]={0,-2,-2,-1,-1,1,1,2,2};int ox[]={0,-1,1,-2,2,-2,2,-1,1};char str[130];bool dfs(int X,int Y,int cnt){ if(cnt==goal) return true; for(int i=0;i<9;i++){ int x=X+ox[i]; int y=Y+oy[i]; if(0<x&&x<=W&&0<y&&y<=H&&!vis[x][y]){ vis[x][y]=true; if(dfs(x,y,cnt+1)){ str[2*cnt]='A'+y-1; str[2*cnt+1]='0'+x; return true; } vis[x][y]=false; //加上這個條件32ms 立馬變0ms if(cnt==0) return false; } } return false;}int main(){ int T; scanf("%d",&T); for(int t=1;t<=T;t++){ memset(vis,false,sizeof(vis)); scanf("%d%d",&W,&H);goal=W*H; printf("Scenario #%d:/n",t); if(dfs(1,1,0)){ str[W*H*2]=0; puts(str); } else puts("impossible"); putchar('/n'); } return 0;}新聞熱點
疑難解答
