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AtCoder Beginner Contest 311 G One More Grid Task

发布时间 2023-07-24 08:09:33作者: zltzlt

洛谷传送门

AtCoder 传送门

考虑一维怎么做。

因为 \(a_i \ge 1\)，所以保持最小值不变的前提下，区间左右端点扩展是更优的。我们使用单调栈求出左边第一个比 \(a_i\) 大的数 \(a_{l_i}\)，以及右边第一个比 \(a_i\) 大的数 \(a_{r_i}\)，答案即为 \(\max\limits_{i = 1}^n a_i \times (\sum\limits_{j = l_i + 1}^{r_i - 1} a_j)\)。

二维的情况，我们枚举矩形的上下边界 \([x, y]\)，令 \(b_i = \min\limits_{j = x}^y a_{j, i}\)，就转化成了一维的情况。时间复杂度 \(O(n^3)\)。

code 
// Problem: G - One More Grid Task
// Contest: AtCoder - Toyota Programming Contest 2023#4（AtCoder Beginner Contest 311）
// URL: https://atcoder.jp/contests/abc311/tasks/abc311_g
// Memory Limit: 1024 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 310;

int n, m, a[maxn][maxn], b[maxn], stk[maxn], top, L[maxn], R[maxn], c[maxn][maxn];

inline int qsum(int xl, int xr, int yl, int yr) {
	return c[xl - 1][yl - 1] + c[xr][yr] - c[xl - 1][yr] - c[xr][yl - 1];
}

void solve() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			scanf("%d", &a[i][j]);
			c[i][j] = c[i - 1][j] + c[i][j - 1] - c[i - 1][j - 1] + a[i][j];
		}
	}
	ll ans = 0;
	for (int i = 1; i <= n; ++i) {
		for (int k = 1; k <= m; ++k) {
			b[k] = a[i][k];
		}
		for (int j = i; j <= n; ++j) {
			for (int k = 1; k <= m; ++k) {
				b[k] = min(b[k], a[j][k]);
			}
			top = 0;
			for (int k = 1; k <= m; ++k) {
				while (top && b[stk[top]] > b[k]) {
					--top;
				}
				L[k] = (top ? stk[top] + 1 : 1);
				stk[++top] = k;
			}
			top = 0;
			for (int k = m; k; --k) {
				while (top && b[stk[top]] >= b[k]) {
					--top;
				}
				R[k] = (top ? stk[top] - 1 : m);
				stk[++top] = k;
			}
			for (int k = 1; k <= m; ++k) {
				ans = max(ans, 1LL * b[k] * qsum(i, j, L[k], R[k]));
			}
		}
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}