【BZOJ】【P1047】【HAOI2007】【理想的正方形】【题解】【单调队列】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1047

先用单调队列横着预处理出每行连续的k个的极值

再竖着做一遍单调队列就行了

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cctype>
#include<climits>
#include<queue>
#include<cstdlib>
using namespace std;
const int maxn=1005;
typedef pair<int,int> pii;
int n,m,k,ans=INT_MAX;
int mp[maxn][maxn];
int Min[maxn][maxn];
int Max[maxn][maxn];
int getint(){
	int res=0;char c=getchar();
	while(!isdigit(c))c=getchar();
	while(isdigit(c))res=res*10+c-'0',c=getchar();
	return res;
}
deque<pii>qmax,qmin;
void deb(deque<pii> q){
	for(int i=0;i<q.size();i++)printf("%d%c",q[i].first," \n"[i==q.size()-1]);
}
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)mp[i][j]=getint();
	for(int i=1;i<=n;i++){
		qmax.clear();qmin.clear();
		for(int j=1;j<=m;j++){
			while(qmax.size()&&qmax.front().second<j-k+1)qmax.pop_front();
			while(qmin.size()&&qmin.front().second<j-k+1)qmin.pop_front();
			while(qmax.size()&&qmax.back().first<=mp[i][j])qmax.pop_back();
			qmax.push_back(pii(mp[i][j],j));
			while(qmin.size()&&qmin.back().first>=mp[i][j])qmin.pop_back();
			qmin.push_back(pii(mp[i][j],j));			
			Min[i][j]=qmin.front().first;Max[i][j]=qmax.front().first;
		}
	}
	for(int j=k;j<=m;j++){
		qmax.clear();qmin.clear();
		for(int i=1;i<=n;i++){
			while(qmax.size()&&qmax.front().second<i-k+1)qmax.pop_front();
			while(qmin.size()&&qmin.front().second<i-k+1)qmin.pop_front();
			while(qmax.size()&&qmax.back().first<=Max[i][j])qmax.pop_back();
			qmax.push_back(pii(Max[i][j],i));
			while(qmin.size()&&qmin.back().first>=Min[i][j])qmin.pop_back();
			qmin.push_back(pii(Min[i][j],i));	
			if(i>=k&&j>=k)
			ans=min(ans,qmax.front().first-qmin.front().first);				
		}
	}cout<<ans<<endl;
	return 0;
}