bzoj1850: Submatrix

传送门
我们考虑枚举大矩形的右上角，预处理出每部最小的矩形
首先我们可以O（N^2）算出以定点为左上角的小矩形的数值和
然后单调队列求出定点在某个区间内的最小和
然后枚举大矩形。
时间复杂度O（N^2)

#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#define fo(i,j,k) for (int i=j;i<=k;i++)#define N 1005using namespace std;int n,m,A,B,C,D,h,t,val,ans;int a[N][N],b[N][N],c[N][N],d[N][N],q[N];int main(){    scanf("%d%d%d%d%d%d",&n,&m,&A,&B,&C,&D);    fo(i,1,n) fo(j,1,m)        scanf("%d",&a[i][j]);    fo(i,1,n) fo(j,1,m)        a[i][j]+=a[i-1][j];    fo(i,1,n) fo(j,1,m)        a[i][j]+=a[i][j-1];    fo(i,C,n) fo(j,D,m)        b[i][j]=a[i][j]-a[i-C][j]-a[i][j-D]+a[i-C][j-D];    fo(i,1,n){        h=t=0;        fo(j,1,m){            while (h!=t&&j-q[h+1]>B-D-2) h++;            while (h!=t&&b[i][q[t]]>b[i][j]) t--;            q[++t]=j;            c[i][j]=b[i][q[h+1]];        }    }    fo(j,1,m){        h=t=0;        fo(i,1,n){            while (h!=t&&i-q[h+1]>A-C-2) h++;            while (h!=t&&c[q[t]][j]>c[i][j]) t--;            q[++t]=i;            d[i][j]=c[q[h+1]][j];        }    }    fo(i,A,n) fo(j,B,m){        val=a[i][j]-a[i-A][j]-a[i][j-B]+a[i-A][j-B];        ans=max(ans,val-d[i-1][j-1]);    }    printf("%d\n",ans);}