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<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
思路:
对每个点对它上面和右边的点进行匹配,如果匹配量大于右上角记录的矩阵最大的值,那么这个值就+1,否则就等于匹配量
代码:
#include<iostream>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
char a[maxn][maxn];
int main()
{
int n,ans;
while(cin>>n,n)
{
ans=1;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>a[i][j];
for(int i=0;i<n;i++)
for(int j=n-1;j>=0;j--)
{
dp[i][j]=1;
if(i==0||j==n-1) continue;
int k=dp[i-1][j+1];
for(int s=1;s<=k;s++)
{
if(a[i-s][j]==a[i][j+s]) dp[i][j]++;
else break;
}
ans=max(ans,dp[i][j]);
}
cout<<ans<<endl;
}
}