个人认为这个代码写的比较清晰, 希望能帮助到大家^^

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef pair<int, int > PII;
const int NUMBER = 510, INF = 0x3f3f3f3f;
const int dx1[4] = {-1, -1, 1, 1}, dy1[4] = {-1, 1, 1, -1};
const int dx2[4] = {-1, -1, 0, 0}, dy2[4] = {-1, 0, 0, -1};
const char map[5] = {'\\', '/', '\\', '/'};

int row, col;
char graph[NUMBER][NUMBER];
int dist[NUMBER][NUMBER];
bool visited[NUMBER][NUMBER];

//双端队列本质是dijkstra算法, 因此需要前置数组初始化
void bfs() {
    memset(dist, 0x3f, sizeof dist);
    memset(visited, false, sizeof visited);

    dist[0][0] = 0;
    PII queue[NUMBER * NUMBER * 2];
    int head = NUMBER * NUMBER, tail = head - 1;
    queue[++tail] = {0, 0};

    while (head <= tail) {
        PII _node = queue[head++];

        if (visited[_node.first][_node.second]) continue;
        visited[_node.first][_node.second] = true;

        for (int i = 0; i < 4; ++i) {
            int new_x = _node.first + dx1[i];
            int new_y = _node.second + dy1[i];

            if (new_x < 0 || new_x > row || new_y < 0 || new_y > col) continue;
//          获取到当前电线块的位置
            int x_pos = _node.first + dx2[i];
            int y_pos = _node.second + dy2[i];

//          计算当前位置的电线需不需要翻转
            int val = graph[x_pos][y_pos] == map[i] ? 0 : 1;

            int new_dist = dist[_node.first][_node.second] + val;
            if (new_dist < dist[new_x][new_y]) {
                dist[new_x][new_y] = new_dist;
                if (val) {
                    queue[++tail] = {new_x, new_y};
                }
                else {
                    queue[--head] = {new_x, new_y};
                }
            }
        }
    }

}

int main() {
    int cases;
    scanf("%d", &cases);

    while (cases--) {
        scanf("%d%d", &row, &col);
        for (int i = 0; i < row; ++i) scanf("%s", graph[i]);

        bfs();

        int res = dist[row][col];
        if (res == INF) puts("NO SOLUTION");
        else printf("%d\n", res);
    }

    return 0;
}