HDU 4618 Palindrome Sub-Array 二分(奇偶)+hash判断是否对称

转自：http://blog.csdn.net/tri_integral/article/details/9472437

有一个全由数字组成的矩阵，你要找到一个最大的子正方形，且这个正方形每一行每一列都是回文串。 二分(奇偶)+hash判断是否对称

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
long long hash_r[2][310][310];
long long hash_c[2][310][310];
long long p[310];
int a[310][310];

void hash(){
int i,j;
for(i=1;i<=n;i++){
for(j=1;j<=m;j++)
hash_r[0][i][j]=hash_r[0][i][j-1]*131+a[i][j];
for(j=m;j>=1;j--)
hash_r[1][i][j]=hash_r[1][i][j+1]*131+a[i][j];
}
for(i=1;i<=m;i++){
for(j=1;j<=n;j++)
hash_c[0][j][i]=hash_c[0][j-1][i]*131+a[j][i];
for(j=n;j>=1;j--)
hash_c[1][j][i]=hash_c[1][j+1][i]*131+a[j][i];
}
}
long long cal_r0(int r,int x,int y){return hash_r[0][r][y]-hash_r[0][r][x-1]*p[y-x+1];}
long long cal_r1(int r,int x,int y){return hash_r[1][r][x]-hash_r[1][r][y+1]*p[y-x+1];}
long long cal_c0(int c,int x,int y){return hash_c[0][y][c]-hash_c[0][x-1][c]*p[y-x+1];}
long long cal_c1(int c,int x,int y){return hash_c[1][x][c]-hash_c[1][y+1][c]*p[y-x+1];}
int check(int l){
int i,j,k;
for(i=1;i+l-1<=n;i++){
for(j=1;j+l-1<=m;j++){
int flag=1;
for(k=i;k<=i+l-1&&flag;k++)
if(cal_r0(k,j,j+l/2-1)!=cal_r1(k,j+l/2+l%2,j+l-1))flag=0;
for(k=j;k<=j+l-1&&flag;k++)
if(cal_c0(k,i,i+l/2-1)!=cal_c1(k,i+l/2+l%2,i+l-1))flag=0;
if(flag)return 1;
}
}
return 0;
}
int main(){
int T,i,j;
scanf("%d",&T);
p[0]=1;
for(i=1;i<310;i++)
p[i]=p[i-1]*131;
while(T--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&a[i][j]);
hash();
int l=0,r=n/2,ans=1;
while (l<=r){
int mid=(l+r)>>1;
if (check(2*mid+1)){
ans=2*mid+1;l=mid+1;
}
else r=mid-1;
}
l=(ans+1)/2,r=n/2;
while (l<=r){
int mid=(l+r)>>1;
if (check(2*mid)){
ans=2*mid;l=mid+1;
}
else r=mid-1;
}
printf("%d\n",ans);
}
return 0;
}