题意
Sol
感觉这个思路还是不错的
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 501, SS = 5e6 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M, Q, ans[SS];
char s[MAXN][MAXN];
bitset<MAXN> f[MAXN][MAXN], g[MAXN][MAXN], Empty;
struct Query {
int x1, y1, x2, y2, id;
}q[SS];
void solve(int l, int r, vector<Query> q) {
if(l > r) return ;
vector<Query> lq, rq;
int mid = l + r >> 1;
//f[i][j]从i,j能到达的mid中的点集
//g[i][j]从mid能到达i, j的点集
for(int i = mid; i >= l; i--) {
for(int j = M; j >= 1; j--) {
f[i][j] = Empty;
if(i == mid) f[i][j][j] = (s[i][j] == '.');
if(i + 1 <= mid && s[i + 1][j] == '.') f[i][j] = f[i][j] | f[i + 1][j];
if(j + 1 <= M && s[i][j + 1] == '.') f[i][j] = f[i][j] | f[i][j + 1];
}
}
for(int i = mid; i <= r; i++) {
for(int j = 1; j <= M; j++) {
g[i][j] = Empty;
if(i == mid) g[i][j][j] = (s[i][j] == '.');
if(i - 1 >= mid && s[i - 1][j] == '.') g[i][j] = g[i][j] | g[i - 1][j];
if(j - 1 >= 1 && s[i][j - 1] == '.') g[i][j] = g[i][j] | g[i][j - 1];
}
}
for(auto &cur : q) {
if(cur.x2 < mid) lq.push_back(cur);
else if(cur.x1 > mid) rq.push_back(cur);
else {
//cout << f[cur.x1][cur.y1] << endl;
//cout << g[cur.x2][cur.y2] << endl;
ans[cur.id] = (f[cur.x1][cur.y1] & g[cur.x2][cur.y2]).count();
}
}
solve(l, mid - 1, lq);
solve(mid + 1, r, rq);
}
int main() {
N = read(); M = read();
for(int i = 1; i <= N; i++) scanf("%s", s[i] + 1);
Q = read();
vector<Query> now;
for(int i = 1; i <= Q; i++) {
q[i].x1 = read(), q[i].y1 = read(), q[i].x2 = read(), q[i].y2 = read(); q[i].id = i;
now.push_back(q[i]);
}
solve(1, N, now);
for(int i = 1; i <= Q; i++) puts(ans[i] ? "Yes" : "No");
return 0;
}
/*
3 3
...
.##
.#.
1
1 1 3 1
3 3
...
.##
.#.
5
1 1 3 3
1 1 1 3
1 1 3 1
1 1 1 2
1 1 2 1
*/