CF598D DFS+连通块

容易发现同一连通块的答案相等 ，故可以一次遍历图中empty点的答案并保存，保存方法可以建立一个数组，下表是对应的连通块序号

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
#define read(i) scanf("%d",&i)
const int maxn = 1e6 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;


int tot;
int cnt=1;    
int n, m;
int vis[1005][1005];
char mp[1005][1005];
int res[1000000 + 5];
int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };

int in(int x, int y) {
    return x > 0 && x <= n && y > 0 && y <= m;
}

void dfs(int x, int y) {
    vis[x][y] = cnt;
    for (int i = 0; i < 4; i++) {
        int xx = x + dx[i], yy = y + dy[i];
        if (in(xx, yy) && !vis[xx][yy]) {
            //printf("%d %d", xx, yy);
            if (mp[xx][yy] == '.')   dfs(xx, yy);
            else if (mp[xx][yy] == '*')  tot++;
        }
    }
}

int main() {
    int k;
    read(n), read(m), read(k);
    for (int i = 1; i <= n; i++) {
        scanf("%s", mp[i] + 1);
    }
    int x, y;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (mp[i][j] == '.' && !vis[i][j]) {
                tot = 0;
                dfs(i, j);
                res[++cnt] = tot;
            }
        }
    }

    while (k--) {
        read(x), read(y);
        printf("%d\n",res[vis[x][y]+1]);
    }
    return 0;
}