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快速切题 poj2488 A Knight's Journey
A Knight‘s Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 31195 | Accepted: 10668 |
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.
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
实际用时 20min
情况:CCCCA 注意java和胡乱改动
注意点:1 组间空行但最后一组没有 2 字典序
#include <cstdio>#include <cstring>using namespace std;int n,m;typedef unsigned long long ull;bool used[8][8];char heap[64][3];const int dx[8]={-2,-2,-1,-1,1,1,2,2},dy[8]={-1,1,-2,2,-2,2,-1,1};bool judge(int x,int y){ if(x>=0&&x<n&&y>=0&&y<m)return true; return false;}bool dfs(int x,int y,int cnt){ used[x][y]=true; heap[cnt][0]=x+‘A‘; heap[cnt++][1]=y+‘1‘; if(cnt==n*m)return true; for(int i=0;i<8;i++){ int tx=x+dx[i],ty=y+dy[i]; if(judge(tx,ty)&&!used[tx][ty]){ if(dfs(tx,ty,cnt))return true; } } used[x][y]=false; return false;}int main(){ int T;scanf("%d",&T); for(int ti=1;ti<=T;ti++){ scanf("%d%d",&m,&n); memset(used,0,sizeof(used)); bool fl=dfs(0,0,0); printf("Scenario #%d:\n",ti); if(fl){ for(int i=0;i<n*m;i++){ printf("%s",heap[i]); } puts(""); } else { puts("impossible"); } if(ti<T)puts(""); } return 0;}
快速切题 poj2488 A Knight's Journey
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