题目
思路
通过3次二维hash来判断是否有贡献，二分矩阵半径计算贡献，注意奇偶性问题，可以通过2次二分，也可以通过类似马拉车的方式解决（好学生打死不拉车）
code:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
int a[1001][1001],b[1001][1001],c[1001][1001];
int n,m;
unsigned long long ha[1001][1001],hb[1001][1001],hc[1001][1001],p[1001],p2[1001],ans;
bool check(int x,int y,int l)
{
if (x>n||y>m||x<l||y<l) return 0;
unsigned long long aa,bb,cc;
aa=ha[x][y]-ha[x][y-l]*p[l]-ha[x-l][y]*p2[l]+ha[x-l][y-l]*p[l]*p2[l];
int yy=m-y+l,xx=n-x+l;
bb=hb[xx][y]-hb[xx][y-l]*p[l]-hb[xx-l][y]*p2[l]+hb[xx-l][y-l]*p[l]*p2[l];
cc=hc[x][yy]-hc[x][yy-l]*p[l]-hc[x-l][yy]*p2[l]+hc[x-l][yy-l]*p[l]*p2[l];
return aa==bb&&bb==cc;
}
int main()
{
cin>>n>>m;
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
cin>>a[i][j];
c[i][m-j+1]=a[i][j];
b[n-i+1][j]=a[i][j];
}
}
p2[0]=p[0]=1;
for (int i=1;i<=1000;i++) p[i]=97*p[i-1],p2[i]=13331*p2[i-1];
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
ha[i][j]=ha[i][j-1]*97+a[i][j];
hb[i][j]=hb[i][j-1]*97+b[i][j];
hc[i][j]=hc[i][j-1]*97+c[i][j];
}
}
for (int i=2;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
ha[i][j]=ha[i-1][j]*13331+ha[i][j];
hb[i][j]=hb[i-1][j]*13331+hb[i][j];
hc[i][j]=hc[i-1][j]*13331+hc[i][j];
}
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
{
int l=1,r=n+m,mn=0;
while (l<=r)
{
int mid=(l+r)>>1;
if (check(i+mid,j+mid,mid*2+1)) l=mid+1,mn=max(mn,mid);
else r=mid-1;
}
ans+=mn;
l=1,r=n+m,mn=0;
while (l<=r)
{
int mid=(l+r)>>1;
if (check(i+mid,j+mid,mid*2)) l=mid+1,mn=max(mn,mid);
else r=mid-1;
}
ans+=mn;
}
}
cout<<ans+n*m;
return 0;
}