Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 996 Accepted Submission(s): 468
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
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<n<=1000), followed by n lines which has n character. There won’t be any blank spaces between characters or the end of line. The input file is ended with a 0.
Output
Each test case output one line, the size of the maximum symmetrical sub- matrix.
Sample Input
3
abx
cyb
zca
4
zaba
cbab
abbc
cacq
0
Sample Output
3
3
Source
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;
#define N 1100
#define MOD 1000000007
#define met(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
char G[N][N];
int dp[N][N];
int main()
{
int n;
while(scanf("%d", &n), n)
{
int i, j, i1, j1, Max=1;
met(G, 0);
met(dp, 0);
for(i=0; i<n; i++)
scanf("%s", G[i]);
for(i=0; i<n; i++)
for(j=0; j<n; j++)
{
if(i==0 || j==n-1)
{
dp[i][j] = 1;
continue;
}
i1=i, j1=j;
while(j1<n && i1>=0 && G[i1][j]==G[i][j1] )
i1--, j1++;
if((i-i1)>=dp[i-1][j+1]+1) dp[i][j] = dp[i-1][j+1] + 1;
else dp[i][j] = i-i1;
Max = max(Max, dp[i][j]);
}
printf("%d
", Max);
}
return 0;
}