xsy子矩形

标签：ios   stream   cer   最小值   最小   double   二分答案   namespace   include   

• 考虑一种解题方法，枚举上下边界L,R， 然后二分答案T，我们要判断的是否存在
  \[ \frac{(sum_j - sum_i)}{2 * (R - L + 1 + j - i)} \ge T \]
• 也就是
  \[ (sum_j - 2Tj) - (sumi - 2Ti) \ge 2T(R - L + 1) \]
• 维护前缀最小值即可
• 然后这样是三次方的， 我们考虑random_shuffle一下枚举的序列， 然后查询的时候就先看一下当前的ans + eps是否可行， 不可行的话就不进行二分直接退出
• 这样期望是会进行lnn次的二分， 就赢了

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define M 515
#define ll long long
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
const double eps = 1e-10, inf = 5e13;
double ans = -1e9;
ll sum[M][M], n, tp, ver[M], xl, xr, yl, yr;
pair<int, int> note[M * M];
ll get(int xa, int ya, int xb, int yb) {
    return sum[xb][yb] - sum[xa - 1][yb] - sum[xb][ya - 1] + sum[xa - 1][ya - 1];
}

bool check(double T, int len, int l, int r) {
    double minn = 0, pl = 0;
    double up = T * 2 * len;
    for(int i = 1; i <= n; i++) {
        double ver2 = ver[i] - T * 2 * i;
        if(ver2 - minn >= up) {
            if(T > ans) {
                ans = T;
                xl = l, yl = pl + 1, xr = r, yr = i;
            }
            return true;
        }
        if(ver2 < minn) minn = ver2, pl = i;
    }
    return false;
}

int main() {
    //freopen("rectangle3.in", "r", stdin);
    n = read();
    for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + read();
    for(int i = 1; i <= n; i++) for(int j = i; j <= n; j++) note[++tp] = mmp(i, j);
    for(int i = 1; i <= tp; i++) {
        int x = 1ll * rand() * rand() % i + 1;
        swap(note[x], note[i]);
    }
    for(int i = 1; i <= tp; i++) {
        int l = note[i].first, r = note[i].second;
        for(int j = 1; j <= n; j++) ver[j] = get(l, 1, r, j);
        if(!check(ans + eps, r - l + 1, l, r)) continue;
        double L = ans, R = inf;
        for(int j = 1; j <= 100; j++) {
            double mid = (L + R) / 2.0;
            if(!check(mid, r - l + 1, l, r)) R = mid;
            else L = mid;
        }
    }
//  cerr<< note[tp].first << " " << note[tp].second << "\n";
    printf("%.10lf\n", ans);
    cout << xl << " " << yl << "\n" << xr << " " << yr << "\n";
    return 0;
}

xsy子矩形

标签：ios   stream   cer   最小值   最小   double   二分答案   namespace   include   

原文地址：https://www.cnblogs.com/luoyibujue/p/10713327.html