AcWing_ 188. Warrior cattle_ bfs

AcWing_188. A warrior cow _bfs

a farmer John There are many cows , He wants to trade one of them Don be called The Knight Of cattle .

This cow has a unique superpower , On the farm like Knight Jump like （ It's the familiar way of horse walking in chess ）.

Although this magical cow can't jump on trees and stones , But it can jump freely on the pasture , We use a ranch  x,y  To represent .

This magical cow likes to eat grass like other cows , Here's a map for you , It says The Knight The beginning of , Trees 、 shrub 、 The location of stones and other obstacles , In addition, there is a bundle of grass .

Now your task is , determine The Knight Want to eat grass , At least how many times .

The Knight The location of is  K  To mark , The location of the obstacle is  *  To mark , The position of the grass is  H  To mark .

Here is an example of a map ：

             11 | . . . . . . . . . .
             10 | . . . . * . . . . . 
              9 | . . . . . . . . . . 
              8 | . . . * . * . . . . 
              7 | . . . . . . . * . . 
              6 | . . * . . * . . . H 
              5 | * . . . . . . . . . 
              4 | . . . * . . . * . . 
              3 | . K . . . . . . . . 
              2 | . . . * . . . . . * 
              1 | . . * . . . . * . . 
              0 ----------------------
                                    1 
                0 1 2 3 4 5 6 7 8 9 0 

The Knight You can follow the... In the figure below  A,B,C,D…A,B,C,D…  This path uses  5  Jump to the grass for the first time （ It is possible that the length of other routes is also  5）：

             11 | . . . . . . . . . .
             10 | . . . . * . . . . .
              9 | . . . . . . . . . .
              8 | . . . * . * . . . .
              7 | . . . . . . . * . .
              6 | . . * . . * . . . F<
              5 | * . B . . . . . . .
              4 | . . . * C . . * E .
              3 | .>A . . . . D . . .
              2 | . . . * . . . . . *
              1 | . . * . . . . * . .
              0 ----------------------
                                    1
                0 1 2 3 4 5 6 7 8 9 0

Be careful ：  Data guarantee there must be solutions .

Input format

The first  1  That's ok ： Two Numbers , Number of columns representing the farm  C  And the number of lines  R.

The first  2..R+1  That's ok : Each line is made up of  C  A string of characters , Together to map the ranch .

Output format

An integer , Represents the minimum number of jumps .

Data range

1≤R,C≤150 

sample input ：

10 11
..........
....*.....
..........
...*.*....
.......*..
..*..*...H
*.........
...*...*..
.K........
...*.....*
..*....*..

sample output ：

5

//
#include<bits/stdc++.h>
using namespace std;

const int dx[]={ 1,1,-1,-1,2,2,-2,-2 };
const int dy[]={ 2,-2,2,-2,1,-1,1,-1 };
const int INF=0x3f3f3f3f;
const int N=222;
bool used[N][N];                                            //  Mark 
char s[N][N];
int n,m,ans;

struct A
{
    int x,y,cnt;
    A( int xx=0,int yy=0,int in=0 ):x(xx),y(yy),cnt(in) {}
}tt;

void bfs( int x,int y,int cnt )
{
    int i,tx,ty;
    queue<A> q;
    q.push( A( x,y,cnt ) ); used[x][y]=1;

    while( !q.empty() )
    {                                                       // front()
        tt=q.front(); q.pop();
        if( s[tt.x][tt.y]=='H' ) { ans=min( ans,tt.cnt ); continue; }

        for( i=0;i<8;i++ )
        {
            tx=tt.x+dx[i]; ty=tt.y+dy[i];
            if( tx>=0 && tx<n && ty>=0 && ty<m && 
            used[tx][ty]==0 && s[tx][ty]!='*' )             // tt.cnt
                { used[tx][ty]=1; q.push( A( tx,ty,tt.cnt+1 ) ); }
        }                                                   // A()  Constructors 
    }
}

int main()
{
    int i,j;

    while( cin>>m>>n )
    {
        memset( used,0,sizeof( used ) );
        memset( s,0,sizeof( s ) );

        for( i=0;i<n;i++ ) cin>>s[i];
        for( i=0;i<n;i++ )
            for( j=0;j<m;j++ )
                if( s[i][j]=='K' ) goto out;
    out:
        ans=INF; bfs( i,j,0 );
        cout<<ans<<endl;
    }
    return 0;
}