[CF1195E] OpenStreetMap - 二维单调队列

[CF1195E] OpenStreetMap - 二维单调队列

Description

给定一个 \(n\times m(n,m \le 3000)\) 的矩阵，求出所有大小为 \(a\times b\) 的子矩形中的最小值的和

Solution

二维单调队列

#include <bits/stdc++.h>
using namespace std;

#define int long long

struct MQueue
{
    deque<pair<int, int>> que;

    void Push(int value, int life)
    {
        while (que.size() && que.back().first >= value)
            que.pop_back();
        que.push_back({value, life});
    }

    void Filter(int time)
    {
        while (que.size() && que.front().second < time)
            que.pop_front();
    }

    int Get(int time)
    {
        Filter(time);
        return que.front().first;
    }
};

vector<int> Process(const vector<int> &src, int width)
{
    MQueue que;
    for (int i = 0; i < width; i++)
        que.Push(src[i], i + width - 1);
    vector<int> ans;
    ans.push_back(que.Get(width - 1));
    for (int i = width; i < src.size(); i++)
        que.Push(src[i], i + width - 1), ans.push_back(que.Get(i));
    return ans;
}

signed main()
{
    int n, m, a, b, g0, x, y, z;
    cin >> n >> m >> a >> b >> g0 >> x >> y >> z;

    vector<vector<int>> A(n, vector<int>(m));
    int g = g0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
        {
            A[i][j] = g;
            g = (g * x + y) % z;
        }

    vector<vector<int>> B(n);
    for (int i = 0; i < n; i++)
        B[i] = Process(A[i], b);

    vector<vector<int>> C(B[0].size(), vector<int>(B.size()));
    for (int i = 0; i < C.size(); i++)
        for (int j = 0; j < C[0].size(); j++)
            C[i][j] = B[j][i];

    int ans = 0;
    for (int i = 0; i < C.size(); i++)
    {
        auto tmp = Process(C[i], a);
        for (int j = 0; j < tmp.size(); j++)
        {
            ans += tmp[j];
        }
    }

    cout << ans << endl;
}

[CF1195E] OpenStreetMap - 二维单调队列

原文：https://www.cnblogs.com/mollnn/p/14381492.html