E - OpenStreetMap

Posted 2022-02-09 thusloop

OpenStreetMap
题意：数字矩阵内求 所有a*b 的子矩阵中的最小值 的和
思路：最小值用单调队列维护，像滑动窗口一样

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=2e5+100;
int g[3010*3010];
int h[3100][3100];

signed main()

	int n,m,a,b,x,y,z;
	cin>>n>>m>>a>>b;
	cin>>g[0]>>x>>y>>z;
	for(int i=1; i<=(n+1)*(m+1); i++)g[i]=(g[i-1]*x+y)%z;
	for(int i=1; i<=n; i++)
	
		for(int j=1; j<=m; j++)
		
			h[i][j]=g[(i-1)*m+j-1];
		
	
	int ans=0;
	deque<pii>q2[m+5];
	for(int j=1; j<=m; j++)
	
		for(int i=1; i<a; i++)
		
			while(!q2[j].empty()&&q2[j].back().fi>=h[i][j]) q2[j].pop_back();
			q2[j].push_back(h[i][j],i);
		
	
	for(int i=1; i<=n-a+1; i++)
	
		deque<pii>q;
		for(int j=1; j<=m; j++)
		
			int now=h[i+a-1][j];
			while(!q2[j].empty()&&q2[j].back().fi>=now) q2[j].pop_back();
			q2[j].push_back(now,i+a-1);
		
		for(int j=1; j<=m; j++)
		
			int now=q2[j].front().fi;
			while(!q.empty()&&q.back().fi>=now) q.pop_back();
			q.push_back(now,j);
			if(j>=b)
			
				//cout<<q.front().fi<<" ";
				ans+=q.front().fi;
				if(q.front().se<=j-b+1)q.pop_front();
			
		
		for(int j=1;j<=m;j++)
		
			if(q2[j].front().se<=i)q2[j].pop_front();
		
	
	cout<<ans<<"\\n";