Codeforces 1301E - Nanosoft（二維RMQ+二分）

題意

給一個$n*m$的矩形，每個格子有一個顏色。總共4種顏色。
一個合法的正方體是由左上四分之一爲顏色1，右上四分之一爲顏色2，左下四分之一爲顏色3，右下四分之一爲顏色4構成。
$Q$次查詢，每次查詢子矩陣$(x1,y1,x2,y2)$中最大的合法正方體面積。如果不存在，輸出0.
$n,m\le 500, q\le 3e5$

解題思路

先對於每個點，二分求出以這個點爲合法正方形左上部分的右下角可以獲得的最大合法正方體大小。複雜度$O(nmlog(n))$
然後每次查詢，二分答案。檢查的時候在對應矩形內求最大值是否大於等於檢查值。使用二維RMQ去$O(1)$查詢，預處理複雜度$O(nm*log(nm))$
總複雜度$O(nm*log(nm)+q*log(n))$

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define fors(i, a, b) for(int i = (a); i < (b); ++i)
using namespace std;
int n, m, Q;
const int maxn = 520;
char s[maxn];
int d[4][maxn][maxn];
int a[maxn][maxn];
bool in(int x, int y){if(x < 1 || x > n || y < 1 || y > m) return false; return true;}
bool block(int x1, int y1, int x2, int y2, int o){
    if(!in(x1, y1) || !in(x2, y2)) return false;
    int sum = (x2-x1+1)*(y2-y1+1);
    return sum == d[o][x2][y2]-d[o][x1-1][y2]-d[o][x2][y1-1]+d[o][x1-1][y1-1];
}
bool check(int x, int y, int r){
    int x1 = x-r+1, y1 = y-r+1, x2, y2;
    if(!block(x1, y1, x, y, 0)) return false;
    y1 += r; y2 = y+r; x2 = x;
    if(!block(x1, y1, x2, y2, 1)) return false;
    x1 += r; x2 += r;
    if(!block(x1, y1, x2, y2, 3)) return false;
    y1 -= r; y2 -= r;
    if(!block(x1, y1, x2, y2, 2)) return false;
    return true;
}
int lg2[maxn];
int dp[9][9][maxn][maxn];
int get_max(int x1, int y1, int x2, int y2){
    int k = lg2[x2-x1+1], l = lg2[y2-y1+1];
    int t1 = max(dp[k][l][x1][y1], dp[k][l][x2-(1<<k)+1][y1]);
    int t2 = max(dp[k][l][x1][y2-(1<<l)+1], dp[k][l][x2-(1<<k)+1][y2-(1<<l)+1]);
    return max(t1, t2);
}
bool check(int x1, int y1, int x2, int y2, int lim){
    if(x1 > x2 || y1 > y2) return false;
    return get_max(x1, y1, x2, y2) >= lim;
}
int main()
{
    lg2[0] = -1;
    fors(i, 1, maxn) lg2[i] = lg2[i>>1]+1;
    cin>>n>>m>>Q;
    fors(i, 1, n+1) {
        scanf("%s", s+1);
        fors(j , 1, m+1)
            if(s[j] == 'G') d[1][i][j] = 1;
            else if(s[j] == 'Y') d[2][i][j] = 1;
            else if(s[j] == 'R') d[0][i][j] = 1;
            else if(s[j] == 'B') d[3][i][j] = 1;
    }
    fors(o, 0, 4)
        fors(i, 1, n+1)
            fors(j, 1, m+1)
                d[o][i][j] += d[o][i][j-1]+d[o][i-1][j]-d[o][i-1][j-1];
    fors(i, 1, n+1) fors(j, 1, m+1){
        int l = 1, r = n;
        while(l <= r){
            if(check(i, j, mid)) a[i][j] = mid, l = mid+1;
            else r = mid-1;
        }
    }
    fors(k, 0, 9)
    fors(l, 0, 9)
    fors(i, 1, n+1)
    fors(j, 1, m+1){
            if(k == 0 && l == 0) {dp[k][l][i][j] = a[i][j]; continue;}
            else if(l == 0) {
                if(i+(1<<(k-1)) <= n)
                    dp[k][l][i][j] = max(dp[k-1][l][i][j], dp[k-1][l][i+(1<<(k-1))][j]);
                else dp[k][l][i][j] = dp[k-1][l][i][j];
            }
            else {
                if(j+(1<<(l-1)) <= m)
                    dp[k][l][i][j] = max(dp[k][l-1][i][j], dp[k][l-1][i][j+(1<<(l-1))]);
                else dp[k][l][i][j] = dp[k][l-1][i][j];
            }
        }
    while(Q--){
        int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        int l = 1, r = n;
        int ans = 0;
        while(l <= r){
            if(check(x1+mid-1, y1+mid-1, x2-mid, y2-mid, mid)) ans = mid, l = mid+1;
            else r = mid-1;
        }
        printf("%d\n", 4*ans*ans);
    }
}