BZOJ 1047 [HAOI2007]理想的正方形 单调队列DP

题解：

刚睡醒写的，总是看不清括号。。。调了半天。。

就是横向一个掉队列，纵向b个单调队列，n^2的求出以(i,j)点为右下角的n*n矩形的最大值和最小值，然后n^2的统计就好~

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>

#define N 1111
#define BUG system("pause")

using namespace std;

long long a,b,n;
long long map[N][N];
long long q[N][N],h[N],t[N],s,e,dq[N];
long long mx[N][N],ans,mn[N][N];

inline void read()
{
    scanf("%lld%d%d",&a,&b,&n);
    for(long long i=1;i<=a;i++)
        for(long long j=1;j<=b;j++)
            scanf("%lld",&map[i][j]);
}

inline void getmax()
{
    for(long long i=1;i<=b;i++) h[i]=1,t[i]=0;
    for(long long i=1;i<=a;i++)
    {
        for(long long j=1;j<=b;j++)
        {
            
            while(h[j]<=t[j]&&i-q[j][h[j]]+1>n) h[j]++;
            while(h[j]<=t[j]&&map[q[j][t[j]]][j]<=map[i][j]) t[j]--;
            q[j][++t[j]]=i;    
        }
        e=0; s=1;
        for(long long j=1;j<=b;j++)
        {
            while(s<=e&&j-dq[s]+1>n) s++;
            while(s<=e&&map[q[dq[e]][h[dq[e]]]][dq[e]]<=map[q[j][h[j]]][j]) e--;
            dq[++e]=j;
            mx[i][j]=map[q[dq[s]][h[dq[s]]]][dq[s]];
        }
    }
}

inline void getmin()
{
    for(long long i=1;i<=b;i++) h[i]=1,t[i]=0;
    for(long long i=1;i<=a;i++)
    {
        for(long long j=1;j<=b;j++)
        {
            while(h[j]<=t[j]&&i-q[j][h[j]]+1>n) h[j]++;
            while(h[j]<=t[j]&&map[q[j][t[j]]][j]>=map[i][j]) t[j]--;
            q[j][++t[j]]=i;    
        }
        e=0; s=1;
        for(long long j=1;j<=b;j++)
        {
            while(s<=e&&j-dq[s]+1>n) s++;
            while(s<=e&&map[q[dq[e]][h[dq[e]]]][dq[e]]>=map[q[j][h[j]]][j]) e--;
            dq[++e]=j;
            mn[i][j]=map[q[dq[s]][h[dq[s]]]][dq[s]];
        }
    }
}

inline void go()
{
    getmin();
    getmax();
    ans=1LL<<60;
    for(long long i=n;i<=a;i++)
        for(long long j=n;j<=b;j++)
            ans=min(ans,mx[i][j]-mn[i][j]);
    printf("%lld\n",ans);
}

int main()
{
    read(),go();
    return 0;
}