HUD1372Knight Moves

題目鏈接：點擊打開鏈接

題意：給出“馬”的初始位置和最終位置，求“馬”行走的最少步數。

分析：典型的BFS

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;

int vis[10][10], fx, lx, fy, ly;//初始位置及最終位置
int dx[8] = {1, 1, -1, -1, 2, 2, -2, -2};//可以走的八個方向
int dy[8] = {2, -2, 2, -2, 1, -1, 1, -1};
char a[3], b[3];
struct node{//每到一個位置（x，y）時花費步數num
    int x, y, num;
};

void bfs(int x, int y){
    node t, p;
    queue<node> q;
    t.x = x;
    t.y = y;
    t.num = 0;
    memset(vis, 0 , sizeof(vis));
    q.push(t);
    vis[t.x][t.y] = 1;
    while(!q.empty()){
        t = q.front();
        q.pop();
        if(t.x == lx && t.y == ly){
            printf("To get from %s to %s takes %d knight moves.\n", a, b, t.num);
            return;
        }
        for(int i = 0; i < 8; i++){
            int nx = t.x + dx[i], ny = t.y + dy[i];
            if(nx > 0 && nx < 9 && ny > 0 && ny < 9 && !vis[nx][ny]){
                p.x = nx;
                p.y = ny;
                p.num = t.num + 1;
                q.push(p);
            }
        }
    }
}

int main(){
    while(~scanf("%s%s", a, b)){
        fx = a[0] - 'a' + 1;
        fy = a[1] - '0';
        lx = b[0] - 'a' + 1;
        ly = b[1] - '0';
        bfs(fx, fy);
    }
    return 0;
}