NUC1333 Knight Moves【DFS】

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#NUC1333 #Knight Moves

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本文介绍了一种使用广度优先搜索(BFS)算法解决骑士在国际象棋盘上从一个位置移动到另一个位置所需的最少步数的问题。通过具体示例展示了如何实现这一算法，并提供了完整的C++代码。

Knight Moves

时间限制: 1000ms 内存限制: 65535KB

问题描述

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.

输入描述

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.

输出描述

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

样例输入

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

样例输出

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.

来源

{Ulm Local 1996}

问题分析：（略）

这个问题和《UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves【BFS】》是同一个问题，代码拿过来用就AC了。

程序说明：参见参考链接。

参考链接：UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves【BFS】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves */  
  
#include <iostream>  
#include <cstring>  
#include <queue>  
  
using namespace std;  
  
const int DIRECTSIZE = 8;  
struct direct {  
    int drow;  
    int dcol;  
} direct[DIRECTSIZE] =  
    {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};  
  
const int MAXN = 8;  
char grid[MAXN][MAXN];  
  
struct node {  
    int row;  
    int col;  
    int level;  
};  
  
node start, end2;  
int ans;  
  
void bfs()  
{  
    queue<node> q;  
  
    memset(grid, ' ', sizeof(grid));  
  
    grid[start.row][start.col] = '*';  
  
    ans = 0;  
    q.push(start);  
  
    while(!q.empty()) {  
        node front = q.front();  
        q.pop();  
  
        if(front.row == end2.row && front.col == end2.col) {  
            ans = front.level;  
            break;  
        }  
  
        for(int i=0; i<DIRECTSIZE; i++) {  
            int nextrow = front.row + direct[i].drow;  
            int nextcol = front.col + direct[i].dcol;  
  
            if(0 <= nextrow && nextrow < MAXN && 0 <= nextcol && nextcol < MAXN)  
                if(grid[nextrow][nextcol] == ' ') {  
                    grid[nextrow][nextcol] = '*';  
                    node v;  
                    v.row = nextrow;  
                    v.col = nextcol;  
                    v.level = front.level + 1;  
                    q.push(v);  
                }  
        }  
    }  
}  
  
int main(void)  
{  
    char startc, endc;  
  
    while(cin >> startc >> start.row >> endc >> end2.row) {  
        start.row--;  
        start.col = startc - 'a';  
        start.level = 0;  
        end2.row--;  
        end2.col = endc - 'a';  
  
        bfs();  
  
        printf("To get from %c%d to %c%d takes %d knight moves.\n", startc, start.row+1, endc, end2.row+1, ans);  
    }  
  
    return 0;  
}