只要该地块处于同一个连通块内，那么必然可以相互到达。
那么查询的结果，就是该地块 (x,y) 所在的连通块的大小。

所以使用一个blockNum 来表是连通块的大小

dfs 代码

#include<iostream>

using namespace std;

const int MAXN = 1005;
const int dirs[][2] = {{-1,0},{1,0},{0,1},{0,-1}};

char grid[MAXN][MAXN];

//每个点(x,y)的blockId
int blockIds[MAXN][MAXN];
//block的大小
int blockNum[MAXN * MAXN];
int block;

int n,m;

bool inArea(int x,int y){
    return x >= 1 && x <= n && y >= 1 && y <= n;
}

void dfs(int x,int y){
    for(auto &d : dirs){
        int nx = x + d[0];
        int ny = y + d[1];
        if(inArea(nx,ny) && !blockIds[nx][ny] &&
           grid[x][y] ^ grid[nx][ny]){
            blockIds[nx][ny] = block;
            blockNum[block]++;
            dfs(nx,ny);
        }
    }
}

int main(){
    cin >> n >> m;
    for(int i = 1;i <= n;i++){
        cin >> grid[i] + 1;
    }
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= n;j++){
            if(!blockIds[i][j]){
                block++;
                blockIds[i][j] = block;
                blockNum[block]++;
                dfs(i,j);
            }
        }
    }
    int qx,qy;
    for(int i = 1;i <= m;i++){
        cin >> qx >> qy;
        cout << blockNum[blockIds[qx][qy]] << endl;
    }
    return 0;
}

bfs 代码

#include<iostream>
#include<queue>

using namespace std;

using PII = pair<int,int>;

const int MAXN = 1005;
const int dirs[][2] = {{-1,0},{1,0},{0,1},{0,-1}};

char grid[MAXN][MAXN];

//每个点(x,y)的blockId
int blockIds[MAXN][MAXN];
//block的大小
int blockNum[MAXN * MAXN];
int block;

int n,m;

bool inArea(int x,int y){
    return x >= 1 && x <= n && y >= 1 && y <= n;
}

void bfs(int x,int y){
    queue<PII> q;
    q.push({x,y});
    while(!q.empty()){
        auto cur = q.front();
        q.pop();
        for(auto &d : dirs){
            int nx = cur.first + d[0];
            int ny = cur.second + d[1];
            if(inArea(nx,ny) && !blockIds[nx][ny] &&
                grid[cur.first][cur.second] ^ grid[nx][ny]){
                blockIds[nx][ny] = block;
                blockNum[block]++;
                q.push({nx,ny});
            }
        }
    }

}

int main(){
    cin >> n >> m;
    for(int i = 1;i <= n;i++){
        cin >> grid[i] + 1;
    }
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= n;j++){
            if(!blockIds[i][j]){
                block++;
                blockIds[i][j] = block;
                blockNum[block]++;
                bfs(i,j);
            }
        }
    }
    int qx,qy;
    for(int i = 1;i <= m;i++){
        cin >> qx >> qy;
        cout << blockNum[blockIds[qx][qy]] << endl;
    }
    return 0;
}