A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 23108 Accepted: 7819 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 讀懂了題目這題就會做了,以前的馬在跳的時候都是在線上,這個題是這方格內還真有些不適應,轉化成在線上就好 [cpp] #include <stdio.h> #include <string.h> int status[100][100],s; int vex[]={-2,-2,2,2,-1,-1,1,1}; int vey[]={-1,1,-1,1,-2,2,-2,2}; char path1[1000],prepath1[1000]; int path2[1000],prepath2[1000],top,n,m,key,k; int main() { void dfs(int x,int y); int i,j,t,tem; scanf("%d",&t); tem=1; while(t--) { scanf("%d %d",&n,&m); memset(status,0,sizeof(status)); if(n==1&&m==1) { printf("Scenario #%d:\n",tem); tem++; printf("A1\n"); printf("\n"); continue; } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { top=0;s=1; memset(status,0,sizeof(status)); prepath1[top]='A'+j-1; prepath2[top++]=i; status[i][j]=1; key=0; k=0; dfs(i,j); if(k==1) { break; } } if(j!=m+1) { break; } } printf("Scenario #%d:\n",tem); tem++; if(i!=n+1) { for(i=0;i<=n*m-1;i++) { printf("%c%d",path1[i],path2[i]); } printf("\n"); }else { printf("impossible\n"); } printf("\n"); } return 0; } void dfs(int x,int y) { int i,j,xend,yend,change; for(i=0;i<=7;i++) { xend=x+vex[i]; yend=y+vey[i]; if(xend>=1&&xend<=n&¥d>=1&¥d<=m&&!status[xend][yend]) { s++; prepath1[top]=yend+'A'-1; prepath2[top++]=xend; status[xend][yend]=1; if(s==n*m) { k=1; if(key==0) { for(j=0;j<=n*m-1;j++) { path1[j]=prepath1[j]; path2[j]=prepath2[j]; } key=1; }else { change=-1; for(j=0;j<=n*m-1;j++) { if(path1[j]>prepath1[j]) { change=1; break; }else if(path1[j]<prepath1[j]) { change=0; break; }else { if(path2[j]>prepath2[j]) { change=1; break; }else if(path2[j]<prepath2[j]) { change=0; break; } } } if(change==1) { for(j=0;j<=n*m-1;j++) { path1[j]=prepath1[j]; path2[j]=prepath2[j]; } } } }else { dfs(xend,yend); } s--; top--; status[xend][yend]=0; } } }