蓝桥杯真题讲解：子矩阵（二维滑动窗口）
一、视频讲解
二、正解代码
cpp
//二维单调队列
#include<bits/stdc++.h>
#define endl '\n'
#define deb(x) cout << #x << " = " << x << '\n';
#define INF 0x3f3f3f3f
#define int long long
using namespace std;
const int mod = 998244353;
const int N = 1e3 + 10;
int g[N][N];
int q[N];
int line_max[N][N], line_min[N][N];
int maxv[N][N], minv[N][N];
int a, b, n, m;
void solve()
{
cin >> n >> m >> a >> b;
for(int i = 0; i < n; i ++)
for(int j = 0; j < m; j ++)
cin >> g[i][j];
//求出每一行的滑动窗口
for(int i = 0; i < n; i ++){
int h = 0, t = -1;
for(int j = 0; j < m; j ++){
if(h <= t and q[h] < j - b + 1){
h ++;
}
while(h <= t and g[i][q[t]] <= g[i][j])
t--;
q[++t] = j;
if(j - b + 1 >= 0){
line_max[i][j - b + 1] = g[i][q[h]];
}
}
}
//对每一行的滑动窗口求一遍滑动窗口
for(int j = 0; j < m; j ++){
int h = 0, t = -1;
for(int i = 0; i < n; i ++){
if(h <= t and q[h] < i - a + 1){
h ++;
}
while(h <= t and line_max[q[t]][j] <= line_max[i][j])
t --;
q[++t] = i;
if(i - a + 1 >= 0)
maxv[i - a + 1][j] = line_max[q[h]][j];
}
}
//求最小值
//先针对每一行，求出每一行的滑动窗口。
for(int i = 0; i < n; i ++){
int h = 0, t = -1;
for(int j = 0; j < m; j ++){
if(h <= t and q[h] < j - b + 1){
h ++;
}
while(h <= t and g[i][q[t]] >= g[i][j])
t --;
q[++ t] = j;
if(j - b + 1 >= 0)
line_min[i][j - b + 1] = g[i][q[h]];
}
}
//针对每一列的滑动黑窗口，求滑动窗口。
for(int j = 0; j < m; j ++){
int h = 0, t = -1;
for(int i = 0; i < n; i ++){
if(h <= t and q[h] < i - a + 1){
h ++;
}
while(h <= t and line_min[q[t]][j] >= line_min[i][j])
t --;
q[++ t] = i;
if(i - a + 1 >= 0)
minv[i - a + 1][j] = line_min[q[h]][j];
}
}
int ans = 0;
for(int i = 0; i < n; i ++){
for(int j = 0; j < m; j ++){
ans = (ans + maxv[i][j] * minv[i][j] % mod) % mod;
}
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
t = 1;
//cin >> t;
while(t--)
solve();
}