【ACWing】188. The cow of samurai style

Subject address:

https://www.acwing.com/problem/content/190/

Farmer John has a lot of cows, and he wants to trade one of them called The Knight. This cow has a unique superpower, jumping like a Knight on the farm (which is what we are familiar with as a horse in chess). Although this magical cow can't jump on trees and rocks, it can jump on the pasture at will. We use one for the pasture $(x,y)$ . This magical cow likes to eat grass like other cows. Give you a map with the starting position of The Knight, the location of trees, bushes, stones and other obstacles, in addition to a bale of grass. Now your task is to determine how many times the Knight needs to jump at least to eat the grass. The position of The Knight uses$K\; is$ used to mark, the position of the obstacle is marked with *, and the position of grass is marked with H. 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 can follow the $A,B,C,D\dots$ This path uses5 5Jump to the grass$5$ times (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

Note: The data guarantee must have a solution.

Input format:
No. $1$ row: two numbers, indicating the number of columnsCC of thefarm$C$ and number of rows$R$ . Chapter$2...R+1$ line: one for each line byCCA string of$C$ characters, which together describe the map of the ranch.

Output format:
an integer representing the minimum number of jumps.

Data range:
$1\le R,C\le 150$

The idea is BFS. Since the number of steps needs to be recorded, it is necessary to traverse hierarchically during BFS. code show as below:

#include <iostream>
#include <queue>

#define x first
#define y second
using namespace std;

typedef pair<int, int> PII;

const int N = 160;
int C, R;
char a[N][N];
queue<PII> q;
bool st[N][N];

int bfs(PII start, PII end) {
    
    
    int dx[] = {
    
    -2, -1, 1, 2, 2, 1, -1, -2};
    int dy[] = {
    
    1, 2, 2, 1, -1, -2, -2, -1};
    q.push(start);
	
	// 存步数
    int res = 0;
    while (!q.empty()) {
    
    
        res++;
        // 分层遍历需要先存一下队列的size
        int size = q.size();
        for (int i = 0; i < size; i++) {
    
    
            auto t = q.front();
            q.pop();
            for (int j = 0; j < 8; j++) {
    
    
                int nx = t.x + dx[j], ny = t.y + dy[j];
                if (0 <= nx && nx < R && 0 <= ny && ny < C && !st[nx][ny] && a[nx][ny] != '*') {
    
    
                	// 走到终点了，直接返回步数
                    if (nx == end.x && ny == end.y) return res;

                    st[nx][ny] = true;
                    q.push({
    
    nx, ny});
                }
            }
        }
    }
    
    // 表示走不到。题目保证有解，所以这句不会执行
    return -1;
}

int main() {
    
    
    cin >> C >> R;
    PII start, end;
    for (int i = 0; i < R; i++)
        for (int j = 0; j < C; j++) {
    
    
            cin >> a[i][j];
            if (a[i][j] == 'K') start = {
    
    i, j};
            else if (a[i][j] == 'H') end = {
    
    i, j};
        }

    cout << bfs(start, end) << endl;

    return 0;
}

Time and space complexity $O\left(RC\right)$ .