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;
}
\[时间复杂度 \ O(nm\log{\max{(n,m)}})
\]
