Given a rectangle which contains N rows and each of them contains a string length of M.
You must find out a symmetrical isosceles right triangle (with two equal edges and a right angle) with two edges parallel two side of the rectangle. Symmetry means the value should be the same according to the shortest altitude of the triangle. And just output the area of the triangle.
Input
The first line of the input contains an integer T (1 <= T <= 20), which mean there are T test case follow.
For each test case, the first line contains two integer number N and M (1 <= N, M <= 500) which means described above.
And then N lines follow, which contains a string of length M. The string only contains letters or digits.
Output
For each test case, output a single integer that is the answer to the problem described above.
Sample Input
1 4 4 abab dacb adab cabb
Sample Output
6
题目链接：点击查看
题意：给出n*m的矩阵,只包含字母和数字,求最大的对称的等腰三角形的面积
题解：对于每个位置，先预处理出对于每个位置斜着对应奇数对称和偶数对称的最大个数，然后暴力枚举每个位置，以这个位置为直角边，分别求4个方向的最大等腰三角形即可
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
int n,m;
char mp[510][510];
int mpji[510][510][2];
int mpou[510][510][2];
int cul(int x1,int y1,int x2,int y2,int kx1,int ky1,int kx2,int ky2)
{
int ans=0;
// cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
while(x1>0&&x1<=n&&x2>0&&x2<=n&&y1>0&&y1<=m&&y2>0&&y2<=m)
{
// cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
// cout<<mp[x1][y1]<<" "<<mp[x2][y2]<<endl;
if(mp[x1][y1]==mp[x2][y2]) ans+=2;
else break;
x1+=kx1;
y1+=ky1;
x2+=kx2;
y2+=ky2;
}
return ans;
}
int main(){
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%s",mp[i]+1);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
mpji[i][j][0]=cul(i,j,i,j,-1,1,1,-1)-1;
mpji[i][j][1]=cul(i,j,i,j,-1,-1,1,1)-1;
mpou[i][j][0]=cul(i,j,i+1,j-1,-1,1,1,-1);
mpou[i][j][1]=cul(i,j,i+1,j+1,-1,-1,1,1);
// cout<<mpou[i][j][0]<<" "<<mpou[i][j][1]<<endl;
}
}
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
int cnt=0,flag=1;
for(int k=1;;k++)
{
if(k&1)
{
if(i-k/2<=0||j-k/2<=0) break;
if(mpji[i-k/2][j-k/2][0]>=k) cnt+=k;
else break;
}
else
{
if(i-k/2<=0||j-k/2+1<=0) break;
if(mpou[i-k/2][j-k/2+1][0]>=k) cnt+=k;
else break;
}
}
ans=max(ans,cnt);
cnt=0;
for(int k=1;;k++)
{
if(k&1)
{
if(i-k/2<=0||j+k/2>m) break;
if(mpji[i-k/2][j+k/2][1]>=k) cnt+=k;
else break;
}
else
{
if(i-k/2<=0||j+k/2-1>m) break;
if(mpou[i-k/2][j+k/2-1][1]>=k) cnt+=k;
else break;
}
}
ans=max(ans,cnt);
cnt=0;
for(int k=1;;k++)
{
if(k&1)
{
if(i+k/2>n||j-k/2<=0) break;
if(mpji[i+k/2][j-k/2][1]>=k) cnt+=k;
else break;
}
else
{
if(i+k/2-1>n||j-k/2<=0) break;
if(mpou[i+k/2-1][j-k/2][1]>=k) cnt+=k;
else break;
}
}
ans=max(ans,cnt);
cnt=0;
for(int k=1;;k++)
{
if(k&1)
{
if(i+k/2>n||j+k/2>m) break;
if(mpji[i+k/2][j+k/2][0]>=k) cnt+=k;
else break;
}
else
{
// cout<<mpou[i+k/2-1][j+k/2][1]<<endl;
if(i+k/2-1>n||j+k/2>m) break;
if(mpou[i+k/2-1][j+k/2][0]>=k) cnt+=k;
else break;
}
}
ans=max(ans,cnt);
}
}
printf("%d\n",ans);
}
return 0;
}
/*
4 4
cbca
bada
cdbb
accc
*/