hdu 1372 Knight Moves（最经典的bfs）

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2136    Accepted Submission(s): 1354

Problem 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 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

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.

 

现在写一下关于我对队列的心得体会：

1.queue<int> Q; //这个就是建立一个队列，名字叫Q，队列里的所有元素都是int整型；

2.Q.push(a);       //这个就是将一个元素a，push进去排队；

3.b=Q.front();     //这个就是队列Q的排最前第一个的元素的值赋给b；

4.Q.pop();          //这个就是将排第一的元素移出队列，第二个的自然跟上；

5.Q.empty();       //这个就是判断队列Q是否为空，如果空就是1，不空就是0。

 

      这几天一直都在做广搜bfs，首先要说一下这题，个人觉得最经典的题目就是这题---骑士游历。这题给我的启发很大，对付这类的问题有信心了。

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

代码：

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

int xx[8] = {1, 2, 1, 2, -1, -2, -1, -2};
int yy[8] = {2, 1, -2, -1, 2, 1, -2, -1};

char c1, c2;
int b1, b2, x1, x2, y1, y2;
bool ch[10][10];

struct point
{
    int x, y, step;
}n, m;

int main()
{
    int i;
    while(cin>>c1>>b1>>c2>>b2)
    {
        x1 = c1 - 'a' + 1;
        x2 = c2 - 'a' + 1;
        y1 = b1;
        y2 = b2;
        memset(ch, false, sizeof(ch));
        n.x = x1; n.y = y1;
        n.step = 0;
        ch[n.x][n.y] = true;
        queue<point> P;
        P.push(n);
        while(!P.empty())
        {
            m = P.front();
            P.pop();
            if(m.x == x2 && m.y == y2) break;
            for(i = 0; i < 8; i++)
            {
                n.x = m.x + xx[i];
                n.y = m.y + yy[i];
                n.step = m.step + 1;
                if(n.x>0 && n.x<=8 && n.y>0 && n.y<=8 && !ch[n.x][n.y])
                {
                    ch[n.x][n.y] = true;
                    P.push(n);
                }
            }
        }
        printf("To get from %c%d to %c%d takes %d knight moves.\n",
                c1, y1, c2, y2, m.step);
    }

    return 0;
}