Codeforces Round #371 (Div. 1) D - Animals and Puzzle 二维ST表 + 二分

时间:2024-04-25 11:06:11

D - Animals and Puzzle

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std; const int N = + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ; int a[N][N], b[N][N], Log[N], n, m; struct ST2 {
int dp[N][N][][], ty;
void build(int n, int m, int b[N][N], int _ty) {
ty = _ty;
for(int i = ; i <= n; i++)
for(int j = ; j <= m; j++)
dp[i][j][][] = ty * b[i][j];
for(int u = ; u <= Log[n]; u++) {
for(int v = ; v <= Log[m]; v++) {
if(!u && !v) continue;
for(int i = ; i+(<<u)- <= n; i++) {
for(int j = ; j+(<<v)- <= m; j++) {
if(u) dp[i][j][u][v] = max(dp[i][j][u-][v], dp[i+(<<(u-))][j][u-][v]);
else dp[i][j][u][v] = max(dp[i][j][u][v-], dp[i][j+(<<(v-))][u][v-]);
}
}
}
}
}
int query(int x1, int y1, int x2, int y2) {
int k1 = Log[x2-x1+], k2 = Log[y2-y1+];
x2 = x2-(<<k1)+;
y2 = y2-(<<k2)+;
return max(max(dp[x1][y1][k1][k2], dp[x2][y1][k1][k2]),
max(dp[x1][y2][k1][k2], dp[x2][y2][k1][k2]));
}
} rmq; int main() {
for(int i = -(Log[]=-); i < N; i++)
Log[i] = Log[i - ] + ((i & (i - )) == ); scanf("%d%d", &n, &m);
for(int i = ; i <= n; i++)
for(int j = ; j <= m; j++)
scanf("%d", &a[i][j]); for(int i = n; i >= ; i--) {
for(int j = m; j >= ; j--) {
if(!a[i][j]) continue;
b[i][j] = min(b[i+][j+], min(b[i][j+], b[i+][j])) + ;
}
} rmq.build(n, m, b, );
int q; scanf("%d", &q);
while(q--) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int l = , r = min(x2-x1, y2-y1) + , ans = ;
while(l <= r) {
int mid = l + r >> ;
int x3 = x2 - mid + , y3 = y2 - mid + ;
if(rmq.query(x1, y1, x3, y3) >= mid) l = mid + , ans = mid;
else r = mid - ;
}
printf("%d\n", ans);
}
return ;
} /*
*/