子矩阵（十四届蓝桥杯python组A）

这题思路就是用两次单调队列。

下面是TIE的代码。9/10；用了priority_queue做单调队列，而不是手动写的。

（手写单调队列好麻烦~~~，脑子晕死，考试的时候过了9个点也ok，）

#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<unordered_map>
#include<vector>
using namespace std;
const int N = 1E3+10;
const int mod = 998244353;
typedef long long ll;
typedef pair<int,int> pll;

priority_queue<pll> maxn[N];
priority_queue<pll,vector<pll>,greater<pll>> minn[N];
int maxnum[N][N];
int minnum[N][N];

priority_queue<pll> maxl[N];
priority_queue<pll,vector<pll>,greater<pll>> minl[N];


int main (){
	int n,m,a,b;cin>>n>>m>>a>>b;
	for(int i = 1;i<=n;++i){
		for(int j = 1;j<=m;++j){
			int a;cin>>a;
			maxn[i].push({a,j});
			minn[i].push({a,j});
			while(maxn[i].top().second<=j-b){
				maxn[i].pop();
			}
			while(minn[i].top().second<=j-b){
				minn[i].pop();
			}
			maxnum[i][j] = maxn[i].top().first;
			minnum[i][j] = minn[i].top().first;
		}

	}	
	ll hou = 0;
	ll ou = 0;
	for(int j = b;j<=m;++j){
		for(int i = 1;i<=n;++i){
			maxl[j].push({maxnum[i][j],i});
			minl[j].push({minnum[i][j],i});
			if(i>=a){
				while(maxl[j].top().second<=i-a){
					maxl[j].pop();
				}
				while(minl[j].top().second<=i-a){
					minl[j].pop();
				}
				hou = (ll)maxl[j].top().first*(ll)minl[j].top().first%mod;
				ou = (ou+hou)%mod;
				cout<<hou<<endl;
			}
		}
	}
	cout<<ou;
	return 0;
} 

单调队列