A Knight's Journey(搜索题)
2016-04-22 15:56
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A Knight's Journey
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status
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.
he 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
Sample Output
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status
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.
he 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
题意:
按照骑士路线行走是否能将所有的点全部走完(一个小于8*8的棋盘)并按照字典序来进行输出所走的路的坐标先输出列(A,B,C,D,E,F,G)再输出行(1,2,3,4,5,6,7,8);由于要求按照字典许输出路线则按照 int dx[8] = {-1, 1, -2,2, -2, 2, -1, 1};int dy[8] = {-2, -2, -1,-1, 1, 1, 2, 2};//为了使其按照字典序排列且先输出列则按照Y从小到大走即可。(且X也要按照从小到大)(在Y相同的情况下按照象限第三、第二、第四、第一的顺序来进行)代码: #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cstdio> using namespace std; int map[88][88]; int vis[88][88]; int n,m; int flag; int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1}; int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};//为了使其按照字典序排列且
先输出列则按照Y从小到大走即可。(且X也要按照从小到大)(在Y相同的情况下
按照象限第三、第二、第四、第一的顺序来进行) void DFS(int x,int y,int step) { map[step][0]=x; map[step][1]=y; if(step==n*m) { flag=1; return ; } for(int i=0; i<8; i++) { int nx=x+dx[i]; int ny=y+dy[i]; if(nx<1||nx>n||ny<1||ny>m||vis[nx][ny]||flag) continue; else { vis[nx][ny]=1; DFS(nx,ny,step+1); vis[nx][ny]=0; } } } int main() { int i,j; int t; cin>>t; int countt=0; while(t--) { flag=0; scanf("%d%d",&n,&m); memset(vis,0,sizeof(vis)); vis[1][1]=1; DFS(1,1,1); cout<<"Scenario #"<<++countt<<":"<<endl; if(flag) { for(int i=1; i<=n*m; i++) { printf("%c%d",map[i][1]-1+'A',map[i][0]); } } else printf("impossible"); printf("\n"); if(t) printf("\n"); } return 0; }
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