P2216 [HAOI2007] 理想的正方形 题解

题目链接：理想的正方形

比较明显的，我们可以用二维 ST 表解决，具体的二维 ST 表的实现，只需要知道一点：

对于 \(st[i][j][t]=max(i \sim i+2^t,j \sim j+2^t)\)，表示的是如图所示的大正方形范围内的最值，它可以拆成从四个小正方形的左端点走 \(2^{t-1}\) 长的小正方形组成，预处理完直接查极差即可。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}

template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e3 + 10;
constexpr int T = 11;
int stMin[N][N][T];
int stMax[N][N][T];
int a[N][N];
int n, m;
#define R1(x) (x+(1<<t-1))

inline void init()
{
    int k = log2(max(n, m)) + 1;
    forn(i, 1, n)
        forn(j, 1, m)stMax[i][j][0] = stMin[i][j][0] = a[i][j];
    forn(t, 1, k)
    {
        forn(i, 1, n-(1<<t)+1)
        {
            forn(j, 1, m-(1<<t)+1)
            {
                stMax[i][j][t] = max({
                    stMax[i][j][t - 1], stMax[R1(i)][j][t - 1], stMax[i][R1(j)][t - 1], stMax[R1(i)][R1(j)][t - 1]
                });
                stMin[i][j][t] = min({
                    stMin[i][j][t - 1], stMin[R1(i)][j][t - 1], stMin[i][R1(j)][t - 1], stMin[R1(i)][R1(j)][t - 1]
                });
            }
        }
    }
}

#define R2(x) (x+len-(1<<k))

inline int query(const int x, const int y, const int len)
{
    int k = log2(len);
    int mx = max({stMax[x][y][k], stMax[R2(x)][y][k], stMax[x][R2(y)][k], stMax[R2(x)][R2(y)][k]});
    int mi = min({stMin[x][y][k], stMin[R2(x)][y][k], stMin[x][R2(y)][k], stMin[R2(x)][R2(y)][k]});
    return mx - mi;
}

int x;

inline void solve()
{
    cin >> n >> m >> x;
    forn(i, 1, n)
        forn(j, 1, m)cin >> a[i][j];
    init();
    int ans = 1e9 + 7;
    forn(i, x, n)
    {
        forn(j, x, m)
        {
            int L = i - x + 1, R = j - x + 1;
            uMin(ans, query(L, R, x));
        }
    }
    cout << ans;
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度 \ O(nm\log{\max{(n,m)}})
\]