Planting Trees 单调队列

　　题意：  求最大子矩形 满足  子矩形的最大值-最小值 小于m

用单调队列维护  stl 的双端队列会超时

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define see(x) (cerr<<(#x)<<'='<<(x)<<endl)
#define pb push_back
#define inf 0x3f3f3f3f
#define CLR(A,v)  memset(A,v,sizeof A)
typedef pair<int,int>pii;
//////////////////////////////////
const int N=500+10;

int qmax[N],qmin[N];
int mx[N],mi[N],ans,mp[N][N],n,m;

int main()
{
    int cas;cin>>cas;
    while(cas--)
    {
        scanf("%d%d",&n,&m);
        rep(i,1,n)rep(j,1,n)scanf("%d",&mp[i][j]);

        int ans=0;
        for(int u=1;u<=n;u++){
            for(int i=1;i<=n;++i)mx[i]=0,mi[i]=inf;

            for(int d=u;d<=n;++d){
                for(int k=1;k<=n;k++)mi[k]=min(mi[k],mp[d][k]),mx[k]=max(mx[k],mp[d][k]);

                int l1=1,l2=1,r1=0,r2=0;
                for(int r=1,l=1;r<=n;r++)
                {
                    while(l1<=r1&&mx[qmax[r1]]<=mx[r])r1--;
                    while(l2<=r2&&mi[qmin[r2]]>=mi[r])r2--;
                    qmax[++r1]=r;qmin[++r2]=r;

                    while(l<=r&&mx[qmax[l1]]-mi[qmin[l2]]>m)
                    {
                        while(l1<=r1&&qmax[l1]<=l)l1++;
                        while(l2<=r2&&qmin[l2]<=l)l2++;
                        l++;

                    }
                    ans=max(ans,(r-l+1)*(d-u+1));
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}