HDU - 1372 - Knight Moves（bfs）

版权声明：Andy https://blog.csdn.net/Alibaba_lhl/article/details/82706748

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::Click here 

思路

给你骑士的出发位置和目标位置，求到达终点所用的最少的步数。其实就是一个模板题啦，不知道暑期集训时为啥没写对啊！

主要是理解骑士的移动规则就可以了，和中国象棋的”马走日“是一样的，然后写一个方向数组，其他的就没差了。

                                                                                                                                                                                              

AC Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 10;

int sx, sy, ex,ey;
char ch1[5], ch2[5];
bool vis[MAXN][MAXN];
int dp[8][2] = {1,-2,-1,-2,2,-1,-2,-1,2,1,-2,1,1,2,-1,2};
struct node
{
    int x, y;
    int step;
    friend bool operator < (node a, node b)
    {
        return a.step > b.step;
    }
};
void bfs(int x1, int y1, int x2, int y2)
{
    int ans = 0;
    node st1, st2;
    memset(vis,false,sizeof(vis));
    st1.x = x1; st1.y = y1; st1.step = 0;
    vis[st1.x][st1.y] = true;
    priority_queue<node> qu;
    qu.push(st1);
    while(!qu.empty())
    {
        st1 = qu.top();
        qu.pop();
        if(st1.x==x2 && st1.y==y2)
        {
            ans = st1.step;
            break;
        }
        for(int i=0; i<8; i++)
        {
            st2.x = st1.x + dp[i][0];
            st2.y = st1.y + dp[i][1];
            if(!vis[st2.x][st2.y] && st2.x<=8 && st2.x>=1 && st2.y<=8 && st2.y>=1)
            {
                st2.step = st1.step + 1;
                vis[st2.x][st2.y] = true;
                qu.push(st2);
            }
        }
    }
    printf("To get from %s to %s takes %d knight moves.\n",ch1,ch2,ans);
}
int main()
{
    while(~scanf("%s %s",ch1,ch2))
    {
        sx = ch1[1] - '0';
        sy = ch1[0] - 'a' + 1;
        ex = ch2[1] - '0';
        ey = ch2[0] - 'a' + 1;
        bfs(sx,sy,ex,ey);
    }
    return 0;
}