UVA - 439 - Knight Moves （BFS）

UVA - 439

Knight Moves

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

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 Specification

The input file 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 Specification

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.

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Rare Topics :: Rare Problems :: Knight
Moves

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 4. Graph :: Breadth First Search :: SSSP
on Unweighted Graph

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures :: Graphs

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 6. Data Structures :: Exercises

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Graph :: Single-Source Shortest Paths (SSSP) :: On
Unweighted Graph: BFS
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思路：题目意思看了半天才懂，就是类似象棋中的马走日，可以朝八个方向走，然后给出两个点，问从一个点到另一点最短步数为多少，可以用BFS来搜索，一步一步模拟过去



AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

char s1[5], s2[5];
int ans;
int vis[10][10];//记忆化搜索 
int dir[8][2] = {{-2, -1},{-2, 1},{-1, 2},{1, 2},{2, 1},{2, -1},{1, -2},{-1, -2}};//朝八个方向 

struct node {
	int x, y, count;		//count用于记录现在是第几层 
	bool operator == (const node &a) const {	//字符重载，用于快速比较 
		return x == a.x && y == a.y;
	} 
};

int bfs(node pos1, node pos2) {		//广搜 
	queue<node> q;
	node pre, cur;
	q.push(pos1);
	vis[pos1.x][pos1.y] = 1;
	while(!q.empty()) {
		pre = q.front();
		q.pop();
		if(pre == pos2) return pre.count;//找到后返回一个值（BFS层数，第几步） 
		for(int i = 0; i < 8; i++) {
			node cur;
			cur.x = pre.x + dir[i][0];
			cur.y = pre.y + dir[i][1];
			if(!vis[cur.x][cur.y] && cur.x >= 1 && cur.x <= 8 && cur.y >= 1 && cur.y <= 8) {
				vis[cur.x][cur.y] = 1;
				cur.count = pre.count + 1;
				q.push(cur);
			}
		}
	}
	return 0;
}

int main() {
	while(scanf("%s %s", s1, s2) != EOF) {
		node pos1, pos2;
		pos1.y = s1[0] - 'a' + 1;
		pos1.x = s1[1] - '0';
		pos1.count = 0;
		pos2.y = s2[0] - 'a' + 1;
		pos2.x = s2[1] - '0';
		pos2.count = 0;
		
		memset(vis, 0, sizeof(vis));
		ans = bfs(pos1, pos2);
		printf("To get from %s to %s takes %d knight moves.\n", s1, s2, ans);
	} 
	return 0;
}