Q

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<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;int dp[1010][1010];char str[1010][1010];int main(){    int n;    while (~scanf("%d", &n) && n)    {        for (int i = 0; i < n; i++)        {            scanf("%s", str[i]);        }        int ans = 1;        bool flag = 0;        for (int i = 0; i < n; i++)        {            //if (flag)                //break;            for (int j = 0; j < n; j++)            {                //if (flag)                    //break;                if (i == 0 || j == n - 1)                {                    dp[i][j] = 1;                    continue;                }                int t1 = i, t2 = j;                while (t1 >= 0 && t2 < n&&str[t1][j] == str[i][t2])                {                    t1--;                    t2++;                }                //检测是否匹配                t1 = i - t1;                if (t1 >= dp[i - 1][j + 1] + 1)                    dp[i][j] = dp[i - 1][j + 1] + 1;                else                {                    dp[i][j] = t1;                    //flag = 1;                }                ans = max(ans, dp[i][j]);            }        }        printf("%d\n", ans);    }    return 0;}