poj2488 A Knight's Journey裸dfs
A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 35868 | Accepted: 12227 |
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
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
Source
搞清楚一个字典序就行了，其余的很简单。这一题的字典序：就是先按列排序，较小的在前。然后按行排序，也是较小的在前。
我的排序是这样的：
int diri[8]={-1,1,-2,2,-2,2,-1,1};
int dirj[8]={-2,-2,-1,-1,1,1,2,2};
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大意很明了，就是找到一个路径让马走完所有的点，不重复不遗漏；思路很容易找到，直接用DFS搜索标记并回溯，一个点一个点作为起点去试；找到后停止；
#include<stdio.h> int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1}; //记录方向 int g,a,b;//g用来记录是否找到解，找到后不再搜索 int vist[26][26],path[26][2]; void find(int i,int j,int k)//i,j是要走的格子，k记录已经走过的步数 { if(k==a*b)//走完了 { for(int i=0;i<k;i++) printf("%c%d",path[i][0]+'A',path[i][1]+1); printf("\n"); g=1; } else for(int x=0;x<8;x++)//8个方向依次搜索 { int n=i+dir[x][0]; int m=j+dir[x][1]; if(n>=0&&n<b&&m>=0&&m<a&&!vist[n][m]&&!g) { vist[n][m]=1;//标记已走 path[k][0]=n,path[k][1]=m; find(n,m,k+1); vist[n][m]=0;//清除标记 } } } int main() { int n; scanf("%d",&n); for(int m=0;m<n;m++) { g=0; scanf("%d %d",&a,&b); for(int i=0;i<a;i++)//一个点一个点的尝试 for(int j=0;j<b;j++) vist[i][j]=0; vist[0][0]=1; path[0][0]=0,path[0][1]=0; printf("Scenario #%d:\n",m+1); find(0,0,1); if(!g) printf("impossible\n"); printf("\n"); } return 0; }
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; const int maxn=30; bool vis[maxn][maxn]; int path[100][2]; int n,m; int next[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1}; bool flag; void dfs(int x,int y,int step){ if(step==m*n){ flag=true; for(int i=0;i<step;i++){ printf("%c%d",path[i][1]+'A'-1,path[i][0]); } printf("\n"); } else for(int k=0;k<8;k++){ int tx=x+next[k][1]; int ty=y+next[k][0]; if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&!vis[tx][ty]&&!flag){ vis[tx][ty]=true; path[step][0]=tx; path[step][1]=ty; dfs(tx,ty,step+1); vis[tx][ty]=false; } } } int main(){ int t; scanf("%d",&t); int Case=0; while(t--){ Case++; memset(vis,false,sizeof(vis)); memset(path,0,sizeof(path)); scanf("%d%d",&n,&m); flag=false; vis[1][1]=true; path[0][0]=1; path[0][1]=1; printf("Scenario #%d:\n",Case); dfs(1,1,1); if(!flag) printf("impossible\n"); printf("\n"); } return 0; }
大意很明了，就是找到一个路径让马走完所有的点，不重复不遗漏；思路很容易找到，直接用DFS搜索标记并回溯，一个点一个点作为起点去试；找到后停止；
