矩形区域[IOI2019]

#include <bits/stdc++.h>
#define N 2520
using namespace std;

template <typename T>
inline void read(T &num) {
	T x = 0, f = 1; char ch = getchar();
	for (; ch > '9' || ch < '0'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch <= '9' && ch >= '0'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ '0');
	num = x * f;	 
}

int n, m, a[N][N], stk[N], cnt, top, ans; 
vector<int> ok[N][N]; //ok[l][r]: a[l][i]~a[r][i]为合法段的i的集合 
pair<int, int> tmp[N]; //临时存答案 
int lok[N][N], rok[N][N]; 

void solve(int *num, int len) { //找出所有合法段 
	cnt = top = 0;
	for (int i = 1; i <= len; i++) {
		while (top && num[i] > num[stk[top]]) {
			if (i > stk[top] + 1) tmp[++cnt] = make_pair(stk[top]+1, i-1); 
			//num[l] < num[r]
			top--;
		}
		if (top) {
			if (i > stk[top] + 1) tmp[++cnt] = make_pair(stk[top]+1, i-1);
			//num[l]>num[r]
			if (num[i] == num[stk[top]]) top--;
			//特殊处理相等情况 
		}
		stk[++top] = i; 
	}
}

void calc(int l, int r, int u, int d) { 
	//左右边界为l,r;上下边界在[u,d]内 
	int len = 0;
	for (int i = u - 1; i <= d + 1; i++) {
		a[0][++len] = a[i][r];
	}
	solve(a[0], len); //找出列的合法段 
	for (int i = 1; i <= cnt; i++) {
		int tl = tmp[i].first + u - 2, tr = tmp[i].second + u - 2;
		if (lok[tl][tr] <= l) ans++;
	}
} 

int main() {
	read(n); read(m);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			read(a[i][j]);
		}
	}	
	for (int i = 2; i < n; i++) { //预处理ok集合 
		solve(a[i], m);
		for (int j = 1; j <= cnt; j++) ok[tmp[j].first][tmp[j].second].push_back(i); 
	}
	for (int r = 2; r < m; r++)  { //枚举矩形右边界 
		for (int i = 1; i <= n; i++) a[0][i] = a[i][r];
		solve(a[0], n);
		for (int i = 1; i <= cnt; i++) {
			if (rok[tmp[i].first][tmp[i].second] + 1 < r)
				lok[tmp[i].first][tmp[i].second] = r;
			rok[tmp[i].first][tmp[i].second] = r;
			//lok[u][d]表示只考虑列的限制，当矩形的上下边界为u,d，右边界为r时，左边界最左是哪里 
		} 
		for (int l = 2; l <= r; l++) { //枚举矩形左边界 
			if (!ok[l][r].size()) continue;
			int lst = ok[l][r][0];
			for (int i = 1; i < ok[l][r].size(); i++) {
				if (ok[l][r][i] > ok[l][r][i-1] + 1) {
					//找出ok[l][r]中的一个连续段 
					calc(l, r, lst, ok[l][r][i-1]);
					lst = ok[l][r][i];
				}
			} 
			calc(l, r, lst, ok[l][r][ok[l][r].size()-1]);
		} 
	} 
	printf("%d\n", ans);
	return 0;
}