题目大意：给定一个二维矩阵，求出现次数最多的a*b的子矩阵
二维Hash，只要记住横纵的BASE不能相同就可以，爱怎么搞怎么搞
一开始写的自然溢出 结果OLE 以为是自然溢出被卡掉了于是写了双取模…… 结果还是OLE
最后发现尼玛这题读入坑爹……字符串里有空格不说，满满的不可见字符是咋回事……
记住不要用scanf读入……可以用gets，或者fread，注意要把一开始的回车过滤掉
getchar读进来全是错的 不知道怎么回事……
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 1010
#define BASE1 2591
#define BASE2 2593
#define MOD1 1000000007
#define MOD2 1000000009
using namespace std;
typedef long long ll;
int m,n,a,b,ans_time;
char s[M][M];
ll sum1[M][M],sum2[M][M],ans1,ans2,BASE1_a[3]={0,1,1},BASE2_b[3]={0,1,1};
namespace Hash_Table{
struct list{
ll hash_value1,hash_value2;
int times;
list *next;
list(ll _,ll __,list *___):
hash_value1(_),hash_value2(__),times(0),next(___) {}
}*head[1001001];
inline int& Hash(ll hash1,ll hash2)
{
list *temp;
int x=hash1*hash2%1001001;
for(temp=head[x];temp;temp=temp->next)
if(temp->hash_value1==hash1&&temp->hash_value2==hash2)
return temp->times;
return head[x]=new list(hash1,hash2,head[x]),head[x]->times;
}
}
void Output()
{
int i,j,k,l;
for(i=a;i<=m;i++)
for(j=b;j<=n;j++)
{
ll hash1=((sum1[i][j]
-sum1[i-a][j]*BASE1_a[1]
-sum1[i][j-b]*BASE2_b[1]
+sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]
)%MOD1+MOD1)%MOD1;
ll hash2=((sum2[i][j]
-sum2[i-a][j]*BASE1_a[2]
-sum2[i][j-b]*BASE2_b[2]
+sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]
)%MOD2+MOD2)%MOD2;
if(hash1==ans1&&hash2==ans2)
{
for(k=i-a+1;k<=i;k++,puts(""))
for(l=j-b+1;l<=j;l++)
putchar(s[k][l]);
return ;
}
}
}
void Find_Position()
{
int i,j;
for(i=a;i<=m;i++)
for(j=b;j<=n;j++)
{
ll hash1=((sum1[i][j]
-sum1[i-a][j]*BASE1_a[1]
-sum1[i][j-b]*BASE2_b[1]
+sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]
)%MOD1+MOD1)%MOD1;
ll hash2=((sum2[i][j]
-sum2[i-a][j]*BASE1_a[2]
-sum2[i][j-b]*BASE2_b[2]
+sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]
)%MOD2+MOD2)%MOD2;
if(hash1==ans1&&hash2==ans2)
printf("%d %d\n",i-a+1,j-b+1);
}
}
int main()
{
int i,j;
scanf("%d %d\n",&m,&n);
for(i=1;i<=m;i++)
{
fread(s[i]+1,n+1,1,stdin);
for(j=1;j<=n;j++)
sum1[i][j]=sum2[i][j]=s[i][j];
}
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
sum1[i][j]=(sum1[i][j]+sum1[i-1][j]*BASE1)%MOD1,
sum2[i][j]=(sum2[i][j]+sum2[i-1][j]*BASE1)%MOD2;
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
sum1[i][j]=(sum1[i][j]+sum1[i][j-1]*BASE2)%MOD1,
sum2[i][j]=(sum2[i][j]+sum2[i][j-1]*BASE2)%MOD2;
cin>>a>>b;
cout<<a<<' '<<b<<endl;
for(i=1;i<=a;i++)
{
BASE1_a[1]*=BASE1;BASE1_a[1]%=MOD1;
BASE1_a[2]*=BASE1;BASE1_a[2]%=MOD2;
}
for(j=1;j<=b;j++)
{
BASE2_b[1]*=BASE2;BASE2_b[1]%=MOD1;
BASE2_b[2]*=BASE2;BASE2_b[2]%=MOD2;
}
for(i=a;i<=m;i++)
for(j=b;j<=n;j++)
{
ll hash1=((sum1[i][j]
-sum1[i-a][j]*BASE1_a[1]
-sum1[i][j-b]*BASE2_b[1]
+sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]
)%MOD1+MOD1)%MOD1;
ll hash2=((sum2[i][j]
-sum2[i-a][j]*BASE1_a[2]
-sum2[i][j-b]*BASE2_b[2]
+sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]
)%MOD2+MOD2)%MOD2;
int& val=Hash_Table::Hash(hash1,hash2);
if(++val>ans_time)
ans_time=val,ans1=hash1,ans2=hash2;
}
Output();
cout<<ans_time<<endl;
Find_Position();
return 0;
}
