UVA - 439 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 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.

分析：典型的BFS迷宫对短路问题。首先把棋盘抽象成一个8×8的二维数组（棋子的移动和坐标的选取没有关系），对输入进行规范化，分别是起点和终点；注意棋子有8种移动方式，把每个方向对坐标的改变量分别保存在两个数组中；定义好数据结构之后就可以用BFS进行搜索了。

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
struct Node {
	int r, c;
	Node(int r, int c):r(r),c(c) {}
};
int G[10][10];
const int mr[] = { -2, -1, 1, 2, 2, 1, -1, -2};
const int mc[] = { 1, 2, 2, 1, -1, -2, -2, -1};
int sr, sc, dr, dc;

Node walk(const Node& u, int dir) {
    return Node(u.r+mr[dir], u.c+mc[dir]);
}

int inside(const Node& u) {
	return u.r>=1&&u.r<=8 && u.c>=1&&u.c<=8;
}

int solve() {
	memset(G, -1, sizeof(G));
	G[sr][sc] = 0;
	queue<Node> q;
	q.push(Node(sr, sc));
    while(!q.empty()) {
        Node u = q.front(); q.pop();
        if(u.r==dr && u.c==dc) return G[dr][dc];
        for(int i=0; i<8; i++) {
			Node v = walk(u, i);
			if(G[v.r][v.c]<0 && inside(v)) {
                G[v.r][v.c] = G[u.r][u.c] + 1;
                q.push(v);
			}
        }
	}
}

int main()
{
	char t1[5], t2[5];
	while(~scanf("%s%s",t1,t2)) {
        sr = t1[0] - 'a' + 1; sc = t1[1] - '0';
        dr = t2[0] - 'a' + 1; dc = t2[1] - '0';
		printf("To get from %s to %s takes %d knight moves.\n", t1, t2, solve());
	}
    return 0;
}