[BZOJ1047][HAOI2007]理想的正方形 二维单调队列

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1047

我们对每矩阵的一列维护一个大小为$n$的单调队列，队中元素为矩阵中元素。然后扫描每一行，再次维护一个大小为$n$的单调队列，队中元素为当前列的队列中取出的最值。$O(n^2)$扫过去就可以了。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int INF=<<;
 int inline readint(){
     int Num;char ch;
     while((ch=getchar())<''||ch>'');Num=ch-'';
     while((ch=getchar())>=''&&ch<='') Num=Num*+ch-'';
     return Num;
 }
 int A,B,N;
 int M[][];
 int qmn1[][],qmn2[][],hmn[],tmn[];
 int qmx1[][],qmx2[][],hmx[],tmx[];
 int Qmn1[],Qmn2[],Hmn,Tmn;
 int Qmx1[],Qmx2[],Hmx,Tmx;
 int main(){
     A=readint();
     B=readint();
     N=readint();
     for(int i=;i<=A;i++)
         for(int j=;j<=B;j++)
             M[i][j]=readint();
     for(int i=;i<=B;i++){
         hmx[i]=hmn[i]=;
         tmx[i]=tmn[i]=;
         for(int j=;j<N;j++){
             while(hmx[i]<=tmx[i]&&qmx1[i][tmx[i]]<=M[j][i]) tmx[i]--;
             qmx1[i][++tmx[i]]=M[j][i];
             qmx2[i][tmx[i]]=j;
             while(hmn[i]<=tmn[i]&&qmn1[i][tmn[i]]>=M[j][i]) tmn[i]--;
             qmn1[i][++tmn[i]]=M[j][i];
             qmn2[i][tmn[i]]=j;
         }
     }
     int ans=INF;
     for(int i=N;i<=A;i++){
         for(int j=;j<=B;j++){
             while(hmx[j]<=tmx[j]&&qmx2[j][hmx[j]]+N<=i) hmx[j]++;
             while(hmx[j]<=tmx[j]&&qmx1[j][tmx[j]]<=M[i][j]) tmx[j]--;
             qmx1[j][++tmx[j]]=M[i][j];
             qmx2[j][tmx[j]]=i;
             while(hmn[j]<=tmn[j]&&qmn2[j][hmn[j]]+N<=i) hmn[j]++;
             while(hmn[j]<=tmn[j]&&qmn1[j][tmn[j]]>=M[i][j]) tmn[j]--;
             qmn1[j][++tmn[j]]=M[i][j];
             qmn2[j][tmn[j]]=i;
         }
         Hmx=Hmn=;
         Tmx=Tmn=;
         for(int j=;j<N;j++){
             while(Hmx<=Tmx&&Qmx1[Tmx]<=qmx1[j][hmx[j]]) Tmx--;
             Qmx1[++Tmx]=qmx1[j][hmx[j]];
             Qmx2[Tmx]=j;
             while(Hmn<=Tmn&&Qmn1[Tmn]>=qmn1[j][hmn[j]]) Tmn--;
             Qmn1[++Tmn]=qmn1[j][hmn[j]];
             Qmn2[Tmn]=j;
         }
         for(int j=N;j<=B;j++){
             while(Hmx<=Tmx&&Qmx2[Hmx]+N<=j) Hmx++;
             while(Hmx<=Tmx&&Qmx1[Tmx]<=qmx1[j][hmx[j]]) Tmx--;
             Qmx1[++Tmx]=qmx1[j][hmx[j]];
             Qmx2[Tmx]=j;
             while(Hmn<=Tmn&&Qmn2[Hmn]+N<=j) Hmn++;
             while(Hmn<=Tmn&&Qmn1[Tmn]>=qmn1[j][hmn[j]]) Tmn--;
             Qmn1[++Tmn]=qmn1[j][hmn[j]];
             Qmn2[Tmn]=j;
             ans=min(ans,Qmx1[Hmx]-Qmn1[Hmn]);
         }
     }
     printf("%d\n",ans);
     return ;
 }