hdu1372 BFS求最短路径长度

C - 广搜基础

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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.
代码:

/*
简单BFS，寻找最短路径长度
*/
#include <iostream>
#include <queue>
#include <stack>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <cstring>
#include <cstdio>
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
int m[10][10];//用来标记，避免重复，bfs常见剪枝
int ax,ay,bx,by;
struct nod
{
    int x,y;
    int step;
};
int f[8][2]={{-2,1},{-2,-1},{-1,2},{-1,-2},{1,2},{1,-2},{2,1},{2,-1}};
queue<nod> q;
int ans;
void bfs()
{
    nod t;
    while(!q.empty())
    {
        t=q.front();
        m[t.x][t.y]=1;
        q.pop();
        if(t.x==bx&&t.y==by)
        {
            ans=t.step;
           return;
        }
        for(int i=0;i<8;i++)
        {
            int xx=t.x+f[i][0];
            int yy=t.y+f[i][1];
            nod n;
            n.x=xx,n.y=yy,n.step=t.step+1;
            if(xx>=1&&xx<=8&&yy>=1&&yy<=8&&!m[xx][yy])
                q.push(n);
        }
    }
}
int main()
{
    char a,c;
    int b,d;
    while(scanf("%c%d %c%d",&a,&b,&c,&d)!=EOF)//输入的时候需要注意，如果一个一个输入，中间加空格，也可以用两个字符串格式化输入（空格截止）
    {
        getchar();
        memset(m,0,sizeof(m));
         ax=b,ay=a-'a'+1;
         bx=d,by=c-'a'+1;//sb的我重复定义
        //cout<<ax<<" "<<ay<<" "<<bx<<" "<<by<<endl;
        nod k;
        k.x=ax,k.y=ay,k.step=0;
        q.push(k);
        bfs();
        cout<<"To get from "<<a<<b<<" to "<<c<<d<<" takes "<<ans<<" knight moves."<<endl;
            while(!q.empty())
        q.pop();//对于有多组数据情况，一定要记得清空队列
    }
    return 0;
}