Phalanx
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
abx
cyb
zca
4
zaba
cbab
abbc
cacq
0
Sample Output
3
3
3
/*
题意：给你一个n*n的字符矩阵，让你求出，最大的对称矩阵的边长，对角线是从左下角到右上角 初步思路：递推 ，从右上角向下递推，dp[i][j]表示是以(i,j)为左下角的最大对称矩阵的边长，这样就得到状态转移方程:
从(i,j)开始向上，向右遍历，如果相等的长度>dp[i-1][j+1]的话，dp[i][j]=dp[i-1][j+1]+1,否则的话就是
dp[i-1][j+1]; #错误：还在找错误原因
*/
#include <bits/stdc++.h>
using namespace std;
int n;
char s[][];
int dp[][];
int maxn=;
void init(){
memset(dp,,sizeof dp);
maxn=;
}
int main(){
// freopen("in.txt","r",stdin);
while(scanf("%d",&n)!=EOF&&n){
init();
for(int i=;i<=n;i++){
scanf("%s",s[i]+);
}
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
if(i==||j==n){
dp[i][j]=;
continue;
}
int a=i,b=j;//表示向上向右遍历的长度
while(a>=&&b<=n&&s[a][j]==s[i][b]){
a--;
b++;
}
// cout<<dp[i][j]<<" ";
a=i-a;
if(a>=dp[i-][j+]+){
dp[i][j]=dp[i-][j+]+;
}else{
dp[i][j]=a;
}
maxn=max(dp[i][j],maxn);
}
// cout<<endl;
}
// for(int i=1;i<=n;i++){
// for(int j=1;j<=n;j++){
// cout<<dp[i][j]<<" ";
// }
// cout<<endl;
// }
printf("%d\n",maxn);
}
return ;
}