hdu1372 BFS求最短路径长度

Description

Input

Output

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;
}