1047: [HAOI2007]理想的正方形 - BZOJ

Description

有一个a*b的整数组成的矩阵，现请你从中找出一个n*n的正方形区域，使得该区域所有数中的最大值和最小值的差最小。
Input

第一行为3个整数，分别表示a,b,n的值第二行至第a+1行每行为b个非负整数，表示矩阵中相应位置上的数。每行相邻两数之间用一空格分隔。
Output

仅一个整数，为a*b矩阵中所有“n*n正方形区域中的最大整数和最小整数的差值”的最小值。
Sample Input
5 4 2
1 2 5 6
0 17 16 0
16 17 2 1
2 10 2 1
1 2 2 2
Sample Output
1
问题规模
（1）矩阵中的所有数都不超过1,000,000,000
（2）20%的数据2<=a,b<=100,n<=a,n<=b,n<=10
（3）100%的数据2<=a,b<=1000,n<=a,n<=b,n<=100

看了题解，过了好几天又看这道题，感慨万分，好久不做RMQ果然什么都忘了

就是先做一遍右边的，把右边n格的最大值最小值存到最左边这一格

然后做一遍向下的，把下面n格的最值存到这一格，现在，每一个n*n的正方形的信息都存到左上角了，扫一遍就行了

 const
     maxn=;
 var
     a,b:array[..maxn,..maxn]of longint;
     n,m,k:longint;

 procedure down(var x:longint;y:longint);
 begin
     if x>y then x:=y;
 end;

 procedure up(var x,y:longint);
 begin
     if x<y then x:=y;
 end;

 procedure main;
 var
     i,j,l,s,ans:longint;
 begin
     read(n,m,k);
     for i:= to n do
       for j:= to m do
         begin
           read(a[i,j]);
           b[i,j]:=a[i,j];
         end;
     l:=;
     while l<k do
       begin
         for i:= to n do
           for j:= to m do
             begin
               s:=j+l;
               down(s,m-l+);
               down(s,j+k-l);
               up(a[i,j],a[i,s]);
               down(b[i,j],b[i,s]);
             end;
         l:=l<<;
       end;
     l:=;
     while l<k do
       begin
         for i:= to n do
           for j:= to m do
             begin
               s:=i+l;
               down(s,n-l+);
               down(s,i+k-l);
               up(a[i,j],a[s,j]);
               down(b[i,j],b[s,j]);
             end;
         l:=l<<;
       end;
     ans:=maxlongint;
     for i:= to n-k+ do
       for j:= to m-k+ do
         down(ans,a[i,j]-b[i,j]);
     write(ans);
 end;

 begin
     main;
 end.

结果写完就被鄙视了，z55250825 ORZ，其实可以用单调队列

 const
     maxn=;
 var
     a,b:array[..maxn,..maxn]of longint;
     n,m,k,ans:longint;

 function min(x,y:longint):longint;
 begin
     if x<y then exit(x);
     exit(y);
 end;

 procedure init;
 var
     i,j:longint;
 begin
     read(n,m,k);
     for i:= to n do
       for j:= to m do
         begin
           read(a[i,j]);
           b[i,j]:=a[i,j];
         end;
 end;

 var
     q:array[..maxn,..]of longint;
     head,tail:longint;

 procedure work;
 var
     i,j:longint;
 begin
     for i:= to n do
       begin
         head:=;
         tail:=;
         for j:= to m do
           begin
             while (tail>=head) and (q[tail,]<=a[i,j]) do
               dec(tail);
             inc(tail);
             q[tail,]:=j;
             q[tail,]:=a[i,j];
             while q[head,]<=j-k do
               inc(head);
             a[i,j]:=q[head,];
           end;
       end;
     for j:= to m do
       begin
         head:=;
         tail:=;
         for i:= to n do
           begin
             while (tail>=head) and (q[tail,]<=a[i,j]) do
               dec(tail);
             inc(tail);
             q[tail,]:=i;
             q[tail,]:=a[i,j];
             while q[head,]<=i-k do
               inc(head);
             a[i,j]:=q[head,];
           end;
       end;
     for i:= to n do
       begin
         head:=;
         tail:=;
         for j:= to m do
           begin
             while (tail>=head) and (q[tail,]>=b[i,j]) do
               dec(tail);
             inc(tail);
             q[tail,]:=j;
             q[tail,]:=b[i,j];
             while q[head,]<=j-k do
               inc(head);
             b[i,j]:=q[head,];
           end;
       end;
     for j:= to m do
       begin
         head:=;
         tail:=;
         for i:= to n do
           begin
             while (tail>=head) and (q[tail,]>=b[i,j]) do
               dec(tail);
             inc(tail);
             q[tail,]:=i;
             q[tail,]:=b[i,j];
             while q[head,]<=i-k do
               inc(head);
             b[i,j]:=q[head,];
           end;
       end;
     ans:=maxlongint;
     for i:=k to n do
       for j:=k to m do
         ans:=min(ans,a[i,j]-b[i,j]);
     write(ans);
 end;

 begin
     init;
     work;
 end.

Wikioi上实在是过不了，然后写了一个C++的

 #include<cstdio>
 using namespace std;

 const int maxn=;

 int a[maxn][maxn],b[maxn][maxn],n,m,k,ans;

 int min(int x,int y)
 {
     return(x<y?x:y);
 }

 void init()
 {
     int i,j;
     scanf("%d%d%d",&n,&m,&k);
     for(i=;i<=n;i++)
         for(j=;j<=m;j++)
         scanf("%d",&a[i][j]),b[i][j]=a[i][j];
 }

 int q[maxn][],head,tail;

 void work()
 {
     int i,j;
     for(i=;i<=n;i++)
     {
         head=;tail=;
         for(j=;j<=m;j++)
         {
             while(tail>=head & q[tail][]<=a[i][j])--tail;
             tail++;
             q[tail][]=j;
             q[tail][]=a[i][j];
             while(q[head][]<=j-k)head++;
             a[i][j]=q[head][];
         }
     }
     for(j=;j<=m;j++)
     {
         head=;tail=;
         for(i=;i<=n;i++)
         {
             while(tail>=head & q[tail][]<=a[i][j])tail--;
             tail++;
             q[tail][]=i;
             q[tail][]=a[i][j];
             while(q[head][]<=i-k)head++;
             a[i][j]=q[head][];
         }
     }
     for(i=;i<=n;i++)
     {
         head=;tail=;
         for(j=;j<=m;j++)
         {
             while(tail>=head & q[tail][]>=b[i][j])tail--;
             tail++;
             q[tail][]=j;
             q[tail][]=b[i][j];
             while(q[head][]<=j-k)head++;
             b[i][j]=q[head][];
         }
     }
     for(j=;j<=m;j++)
     {
         head=;tail=;
         for(i=;i<=n;i++)
         {
             while(tail>=head & q[tail][]>=b[i][j])tail--;
             tail++;
             q[tail][]=i;
             q[tail][]=b[i][j];
             while(q[head][]<=i-k)head++;
             b[i][j]=q[head][];
         }
     }
     ans=;
     for(i=k;i<=n;i++)
         for(j=k;j<=m;j++)
         ans=min(ans,a[i][j]-b[i][j]);
     printf("%d",ans);
 }

 int main()
 {
     init();
     work();
     return ;
 }