POJ 2243 Knight Moves(BFS)
2016-04-29 22:21
309 查看
Knight Moves
Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard
exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column
and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
Sample Output
解题思路: 是一道比较基础的搜索问题,只需考虑马的跳跃规则,而不需要考虑蹩马腿的情况,有一点特殊的地方就在于,这道题并没有给出明确的图,所以需要自己构造一张图,当然其实只需要考虑是否被重复走过就可以了。
BFS代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13192 | Accepted: 7395 |
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard
exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column
and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves.
解题思路: 是一道比较基础的搜索问题,只需考虑马的跳跃规则,而不需要考虑蹩马腿的情况,有一点特殊的地方就在于,这道题并没有给出明确的图,所以需要自己构造一张图,当然其实只需要考虑是否被重复走过就可以了。
BFS代码:
#include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; struct point { int x,y; }; //int sx,sy,ex,ey;//s->start y->end point s,e; int step[10][10]; string map[10]; int dx[8] = {1,1,-1,-1,2,2,-2,-2}; int dy[8] = {2,-2,2,-2,1,-1,1,-1}; int BFS(point s) { queue<point>P; //memset(step,0,sizeof(step)); int i; P.push( s ); point hd; step[s.x][s.y] = 0; while(!P.empty()) { hd = P.front(); P.pop(); if(hd.x==e.x && hd.y==e.y) return step[hd.x][hd.y]; for(i=0;i<8;i++) { int x = hd.x + dx[i]; int y = hd.y + dy[i]; if(x>=1 && x<=8 && y>=1 && y<=8 && map[x][y]!='@') { step[x][y] = step[hd.x][hd.y] + 1; map[x][y] = '@'; point t; t.x = x; t.y = y; P.push(t); } } } } int main() { //freopen("test.txt","r",stdin); char s1[5],s2[5]; int i; while(cin>>s1>>s2)//~scanf("%s %s",s1,s2) { s.x = s1[1] - '0'; s.y = s1[0] - 'a' + 1; e.x = s2[1] - '0'; e.y = s2[0] - 'a' + 1; for(i=1;i<=8;i++) map[i] = "........."; printf("To get from %s to %s takes %d knight moves.\n",s1,s2,BFS( s )); } //cout << "Hello world!" << endl; return 0; }
相关文章推荐
- 最大序列和
- 文件操作
- 项目经理多年的经验之谈
- windows通讯端口初始化失败
- 工具类Log
- 使用KMS批量激活操作系统
- 删除ORacle 命名空间
- Dubbo与Zookeeper、SpringMVC整合和使用(负载均衡、容错)
- 冲刺第六天
- 15.oracle的dump理解十五 SQL命令DUMP
- Corosync+Pacemaker+MySQL+DRBD(二) 推荐
- 14.oracle的dump理解十四 窥视内存
- 构建之法阅读笔记01
- 13.oracle的dump理解十二 dump heap
- 后缀数组 POJ 3581 Sequence
- 12.oracle的dump理解十二 dump library_cache_object
- 2.Python操作Redis:列表(List)
- 【图文详解】scrapy爬虫与动态页面——爬取拉勾网职位信息(1)
- 【图文详解】scrapy爬虫与动态页面——爬取拉勾网职位信息(1)
- wordpress必装的插件 wp最常用的十个插件