TM 行最小値問題 (SMAWK 法, in O(H + W))
(Optimization/smawk.hpp)

View this file on GitHub
Last update: 2026-11-12 11:56:29+09:00
Include: #include "Optimization/smawk.hpp"

概要

$H \times W$ 行列 $A$ が Totally Monotone であるとき、計算量 $O(H + W)$ で、$A$ の各行に対する最小値を求める。

仕様

template<class VAL, class FUNC> vector<pair<VAL, int>> SMAWK(int H, int W, const FUNC &f)

$H \times W$ 行列の $(i, j)$ 成分が $f(i, j)$ で表されるとする
返り値: 受け取る変数を res として
- res[l].first: $l$ 行目の最小値
- res[l].second: $l$ 行目の値が最小となる列の番号

問題例

COLOCON 2018 Final C - スペースエクスプローラー高橋君

参考文献

tatyam: Monge の手引き書
週刊 spaghetti_source: Totally Monotone Matrix Searching (SMAWK algorithm)

Required by

Verified with

Code

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

// find min_j f(i, j) for all i, by SMAWK, O(H + W)
// f(i, j) must be totally monotone
// slide access NG
template<class VAL, class FUNC> void SMAWKRec
(const vector<int> &X, const vector<int> &Y, const FUNC &f, vector<pair<VAL, int>> &res) {
    if (X.empty()) return;

    // Reduce Step
    vector<int> X2, Y2;
    for (auto y : Y) {
        while (!Y2.empty()) {
            int py = Y2.back(), x = X[(int)Y2.size() - 1];
            if (f(x, y) >= f(x, py)) break;
            Y2.pop_back();
        }
        if (Y2.size() < X.size()) Y2.emplace_back(y);
    }

    // Recurse Step
    for (int i = 1; i < (int)X.size(); i += 2) X2.emplace_back(X[i]);
    SMAWKRec(X2, Y2, f, res);

    // Interpolate Step
    int p = 0;
    for (int i = 0; i < (int)X.size(); i += 2) {
        int lim = (i + 1 < (int)X.size() ? res[X[i + 1]].second : Y.back()), best = Y[p];
        while (Y[p] < lim) {
            p++;
            if (f(X[i], Y[p]) < f(X[i], best)) best = Y[p];
        }
        res[X[i]] = {f(X[i], best), best};
    }
}
template<class VAL, class FUNC> vector<pair<VAL, int>> SMAWK(int H, int W, const FUNC &f) {
    if (H == 0) return {};
    assert(W > 0);
    vector<pair<VAL, int>> res(H, make_pair(numeric_limits<VAL>::max() / 2, -1));
    vector<int> X(H), Y(W);
    for (int i = 0; i < H; i++) X[i] = i;
    for (int j = 0; j < W; j++) Y[j] = j;
    SMAWKRec(X, Y, f, res);
    return res;
}

#line 2 "Optimization/smawk.hpp"
#include <bits/stdc++.h>
#line 4 "Optimization/smawk.hpp"
using namespace std;

// find min_j f(i, j) for all i, by SMAWK, O(H + W)
// f(i, j) must be totally monotone
// slide access NG
template<class VAL, class FUNC> void SMAWKRec
(const vector<int> &X, const vector<int> &Y, const FUNC &f, vector<pair<VAL, int>> &res) {
    if (X.empty()) return;

    // Reduce Step
    vector<int> X2, Y2;
    for (auto y : Y) {
        while (!Y2.empty()) {
            int py = Y2.back(), x = X[(int)Y2.size() - 1];
            if (f(x, y) >= f(x, py)) break;
            Y2.pop_back();
        }
        if (Y2.size() < X.size()) Y2.emplace_back(y);
    }

    // Recurse Step
    for (int i = 1; i < (int)X.size(); i += 2) X2.emplace_back(X[i]);
    SMAWKRec(X2, Y2, f, res);

    // Interpolate Step
    int p = 0;
    for (int i = 0; i < (int)X.size(); i += 2) {
        int lim = (i + 1 < (int)X.size() ? res[X[i + 1]].second : Y.back()), best = Y[p];
        while (Y[p] < lim) {
            p++;
            if (f(X[i], Y[p]) < f(X[i], best)) best = Y[p];
        }
        res[X[i]] = {f(X[i], best), best};
    }
}
template<class VAL, class FUNC> vector<pair<VAL, int>> SMAWK(int H, int W, const FUNC &f) {
    if (H == 0) return {};
    assert(W > 0);
    vector<pair<VAL, int>> res(H, make_pair(numeric_limits<VAL>::max() / 2, -1));
    vector<int> X(H), Y(W);
    for (int i = 0; i < H; i++) X[i] = i;
    for (int j = 0; j < W; j++) Y[j] = j;
    SMAWKRec(X, Y, f, res);
    return res;
}

TM 行最小値問題 (SMAWK 法, in O(H + W)) (Optimization/smawk.hpp)

概要

仕様

問題例

参考文献

Required by

Verified with

Code

TM 行最小値問題 (SMAWK 法, in O(H + W))
(Optimization/smawk.hpp)