sub1MLE求调

#include<iostream>
using namespace std;
const int N = 1e4 + 5;

int n, g[N][N], d[N][N], vis[N][N], dir[4][2]{{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
int cnt = 0;

void dfs(int x, int y){
    d[x][y] = ++cnt;
    for(int i = 0; i < 4; i++){
        int nx = x + dir[i][0];
        int ny = y + dir[i][1];
        if(nx > 0 && nx <= n && ny > 0 && ny <= n && vis[nx][ny] == 0 && g[x][y] ^ g[nx][ny] == 1){
            vis[nx][ny] = 1;
            dfs(nx, ny);
        }
    }
}
void sfd(int x, int y){
    d[x][y] = cnt;
    for(int i = 0; i < 4; i++){
        int nx = x + dir[i][0];
        int ny = y + dir[i][1];
        if(nx > 0 && nx <= n && ny > 0 && ny <= n && vis[nx][ny] == 1 && g[x][y] ^ g[nx][ny] == 1){
            vis[nx][ny] = 2;
            sfd(nx, ny);
        }
    }
}

int main(){
    int m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            char c;
            cin >> c;
            g[i][j] = c - '0';
        }
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            cnt = 0;
            if(!vis[i][j]){
                vis[i][j] = 1;
                dfs(i, j);
                vis[i][j] = 2;
                sfd(i, j);
            }
        }
    }
    for(int i = 0; i < m; i++){
        int x, y;
        cin >> x >> y;
        cout << d[x][y] << '\n';
    }

    return 0;
}

评测记录https://www.luogu.com.cn/record/163025292