HDU1372&POJ2243Knight Moves(BFS呀,转换一下)_start=q.front(); q.pop(); if(start.x==ex&&start.y=-优快云博客

这篇博客介绍了如何解决HDU1372和POJ2243这两道关于骑士在特殊棋盘上移动的问题。文章指出，尽管题目有所不同，但基本思路是将棋盘进行转换，然后应用广度优先搜索(BFS)算法来求解。作者提供了AC代码作为解决方案。

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Knight Moves

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13145		Accepted: 7369

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

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

Ulm Local 1996

POJ链接:http://poj.org/problem?id=2243

HDU链接:http://acm.hdu.edu.cn/showproblem.php?pid=1372

题意:还是马踏棋盘,和上一篇(http://blog.youkuaiyun.com/hurmishine/article/details/50939508)差不多,只不过这题的棋盘有点"特别"而已,简单转换一下就可以了.

AC代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

int step;
int to[8][2] = {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};//骑士移动的8个方位
int map[10][10],ex,ey;
char s1[5],s2[5];

struct node
{
    int x,y,step;
};

int bfs()
{
    int i;
    queue<node> Q;
    node start,next,mid;
    start.x = s1[0]-'a';
    start.y = s1[1]-'1';
    start.step = 0;
    ex = s2[0]-'a';//end
    ey = s2[1]-'1';
    memset(map,0,sizeof(map));
    map[start.x][start.y] = 1;
    Q.push(start);
    while(!Q.empty())
    {
        mid = Q.front();
        Q.pop();
        if(mid.x == ex && mid.y == ey)
            return mid.step;
        for(i = 0; i<8; i++)
        {
            next.x = mid.x+to[i][0];
            next.y = mid.y+to[i][1];
            if(next.x == ex && next.y == ey)
                return mid.step+1;
            if (next.x>=0&&next.x<8&&next.y>=0&&next.y<8&&map[next.x][next.y]==0)
            {
                next.step=mid.step+1;
                map[next.x][next.y]=1;
                Q.push(next);
            }
        }
    }
    return 0;
}

int main()
{
    while(~scanf("%s%s",s1,s2))
    {
        printf("To get from %s to %s takes %d knight moves.\n",s1,s2,bfs());
    }
    return 0;
}