Hdu 2859 Phalanx

Phalanx

Problem Description
Today is army day, but the servicemen are busy with the phalanx for the celebration of the 60th anniversary of the PRC.
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc

Input
There are several test cases in the input file. Each case starts with an integer n (0

#include<bits/stdc++.h>using namespace std;const int maxn=1005;char ans[maxn][maxn];int dp[maxn][maxn];int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int n;    while(cin>>n&&n) {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++) {            for(int j=1;j<=n;j++) {                cin>>ans[i][j];            }        }        int sum=1;        for(int i=1;i<=n;i++) {            for(int j=1;j<=n;j++) {                int cnt=1;                while(ans[i-cnt][j]==ans[i][j+cnt]) cnt++;                if(cnt>dp[i-1][j+1]) dp[i][j]=dp[i-1][j+1]+1;                else dp[i][j]=cnt;                sum=max(sum,dp[i][j]);                /*for(int x=0;x<=n;x++) {                    for(int y=0;y<=n;y++) {                        cout<<dp[x][y]<<" ";                    }                    cout<<endl;                }                cout<<endl;*/            }        }        cout<<sum<<endl;    }    return 0;}