HDU 2859 Phalanx 求最大的对称子矩阵的大小

Phalanx

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1050    Accepted Submission(s): 506

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<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

3abxcybzca4zabacbababbccacq0

 

Sample Output

33

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

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对于每个点，向它的上和右进行dp，如果全都满足，则为右上方的那个点加1，否则就是右上方那个点的值。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdio>#include<math.h>#include<algorithm>#define esp 1e-8using namespace std;int dp[1005][1005];char s[1005][1005];int main(){    int n;    int i,j;    while(cin>>n,n)    {        for(i=1; i<=n; i++)        {            scanf("%s",s[i]+1);            int l=strlen(s[j]);            for(j=1; j<=n; j++)                dp[i][j]=1;        }        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)            {                int op=dp[i-1][j+1];                int flag=0;                int da;                for(int k=1; k<=op; k++)                {                    if(s[i-k][j]==s[i][j+k])                        continue;                    else                    {                        flag=1;                        da=k;                        break;                    }                }                if(flag)                    dp[i][j]=da;                else                    dp[i][j]=op+1;            }        int ans=0;        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                ans=max(ans,dp[i][j]);        cout<<ans<<endl;    }    return 0;}