poj 2488(dfs)
2014-09-06 13:59
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题目链接:http://poj.org/problem?id=2488
A Knight's Journey
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
Sample Output
Source
TUD Programming Contest 2005, Darmstadt, Germany
思路:刚开始用广搜解题,但是后来要按字典序输出,所以改为深搜~只要注意方向数组的设置就行;
A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 31012 | Accepted: 10613 |
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
思路:刚开始用广搜解题,但是后来要按字典序输出,所以改为深搜~只要注意方向数组的设置就行;
#include <iostream> #include <stdio.h> #include <string.h> #include <string> #include <cstdio> using namespace std; int xx[]={-1,1,-2,2,-2,2,-1,1}; //设置方向数组优化搜索 int yy[]={-2,-2,-1,-1,1,1,2,2}; bool vis[30][30];//标记状态 int n,m,flag; int map[30][30]; //地图 int path[30][30];//存储路径 void dfs(int x,int y,int k) { if(k==n*m) { for(int i=0;i<k;i++) printf("%c%d",64+path[i][0],path[i][1]); flag=1; return; } for(int i=0;i<8;i++) { int dx=x+xx[i]; int dy=y+yy[i]; if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&!vis[dx][dy]&&!flag) { vis[dx][dy]=true; path[k][0]=dy; path[k][1]=dx; dfs(dx,dy,k+1); vis[dx][dy]=false; } } } int main() { int T,test=1; cin>>T; while(T--) { scanf("%d%d",&n,&m); memset(vis,false,sizeof(vis)); path[0][0]=1; path[0][1]=1; vis[1][1]=true; flag=0; printf("Scenario #%d:\n",test++); dfs(1,1,1); if(!flag)printf("impossible\n"); else printf("\n"); printf("\n"); } return 0; }
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