小A与小B

分部扩展小A和小B的移动范围，并判断是否能相遇。

当st[1][x][y] == st[2][x][y]就相遇

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 1e3 + 5;
char g[N][N];
bool st[2][N][N];
queue<pair<int, int>> q[2];

int n, m;
int x, y, x2, y2;

int dx[] = {0, 0, -1, 1, 1, -1, -1, 1};
int dy[] = {1, -1, 0, 0, 1, 1, -1, -1};

bool bfs(int t) {
    int sz = q[t].size();
    while(sz--){
        auto p = q[t].front();
        q[t].pop();
        int x = p.first, y = p.second;

        for(int i = 0; i < 4 + 4 * (!t); ++ i) {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if(nx >= n || nx < 0 || ny >= m || ny < 0) continue;
            if(g[nx][ny] == '#' || st[t][nx][ny]) continue;
            if(st[!t][nx][ny]) return true;
            st[t][nx][ny] = true;
            q[t].push({nx, ny});
        }
    }
    return false;
}


int min_steps() {
    q[0].push({x, y});
    q[1].push({x2, y2});
    st[0][x][y] = true;
    st[1][x2][y2] = true;
    int res = 0;
    while(!q[0].empty()||!q[1].empty()) {
        res ++;
        if(bfs(0)) return res;
        if(bfs(1)) return res;
        if(bfs(1)) return res;
    }
    return -1;
}


int main() {
    cin >> n >> m;
    for(int i = 0; i < n; ++ i)
        for(int j = 0; j < m; ++ j)
        {
            cin >> g[i][j];
            if(g[i][j] == 'C') x = i, y = j;
            if(g[i][j] == 'D') x2 = i, y2 = j;
        }
   int res = min_steps();
   if(res == -1) puts("NO");
   else {
       puts("YES");
       cout  << res << endl;
   }
   return 0;
}