Bzoj1047理想的正方形

题目：Bzoj1047

题解：

首先，对于每行长度为n的区间求出RMQ，再每一列求出先前求出每一行的RMQ的RMQ，这样做相当于求出了边长为n的正方形的RMQ，用正方形的左上角记录，枚举左上角的坐标求解。

求长度为n的区间的RMQ时可以用单调队列，时间复杂的为O(n)。

代码：

program pro;
var
        map:array[0..1005,0..1005]of longint;
        zx,zy:array[0..1005,0..1005]of record max,min:longint; end;
        que,que1,time,time1:array[0..10000]of longint;
        ans,a,b,n,i,j,l,r,h,h1,t,t1,tt:longint;

procedure openf;
begin
        assign(input,'square.in'); reset(input);
        assign(output,'square.out'); rewrite(output);
end;

procedure closef;
begin
        close(input);
        close(output);
end;

begin
        openf;
        readln(a,b,n);
        for i:=1 to a do
        for j:=1 to b do read(map[i,j]);

        for i:=1 to a do
        begin
                h:=1; t:=0;
                h1:=1; t1:=0;
                fillchar(que,sizeof(que),0);
                fillchar(que1,sizeof(que1),0);
                for j:=1 to n do
                begin
                        while (t>=h)and(map[i,j]>que[t]) do dec(t);
                        inc(t); que[t]:=map[i,j]; time[t]:=j;
                        while (t1>=h1)and(map[i,j]<que1[t1]) do dec(t1);
                        inc(t1); que1[t1]:=map[i,j]; time1[t1]:=j;

                end;
                for l:=1 to b-n+1 do
                begin
                        r:=l+n-1;
                        while (t>=h)and(que[t]<map[i,r]) do dec(t);
                        inc(t); que[t]:=map[i,r]; time[t]:=r;
                        while (t1>=h1)and(que1[t1]>map[i,r]) do dec(t1);
                        inc(t1); que1[t1]:=map[i,r]; time1[t1]:=r;
                        while time[h]<l do inc(h);
                        zx[i,l].max:=que[h];
                        while time1[h1]<l do inc(h1);
                        zx[i,l].min:=que1[h1];
                end;
        end;

        for i:=1 to b-n+1 do
        begin
                h:=1; t:=0;
                h1:=1; t1:=0;
                fillchar(que,sizeof(que),0);
                fillchar(que1,sizeof(que1),0);
                for j:=1 to n do
                begin
                        while (t>=h)and(zx[j,i].max>que[t]) do dec(t);
                        inc(t); que[t]:=zx[j,i].max; time[t]:=j;
                        while (t1>=h1)and(zx[j,i].min<que1[t1]) do dec(t1);
                        inc(t1); que1[t1]:=zx[j,i].min; time1[t1]:=j;
                end;
                for l:=1 to a-n+1 do
                begin
                        r:=l+n-1;
                        while (t>=h)and(que[t]<zx[r,i].max) do dec(t);
                        inc(t); que[t]:=zx[r,i].max; time[t]:=r;
                        while (t1>=h1)and(que1[t1]>zx[r,i].min)do dec(t1);
                        inc(t1); que1[t1]:=zx[r,i].min; time1[t1]:=r;
                        while time[h]<l do inc(h);
                        zy[l,i].max:=que[h];
                        while time1[h1]<l do inc(h1);
                        zy[l,i].min:=que1[h1];
                end;
        end;

        ans:=maxlongint;
        for i:=1 to a-n+1 do
        for j:=1 to b-n+1 do
        if zy[i,j].max-zy[i,j].min<ans then ans:=zy[i,j].max-zy[i,j].min;
        writeln(ans);
        closef;

end.