HDU2859(KB12-Q DP)

Phalanx

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1542    Accepted Submission(s): 757

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

3

abx

cyb

zca

4

zaba

cbab

abbc

cacq

0

 

Sample Output

3

3

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

题意：找给出的方阵中，以左下到右上方向为对称轴的最大对称子方阵。

题解：dp[i][j]表示以ch[i][j]格子为左下角的最大对称子方阵。dp[i][j] = min(dp[i-1][j+1]+1, 从(i,j)点向上、右方向匹配的对称数)

 //2017-04-20
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;
 char ch[N][N];
 int dp[N][N];

 int main()
 {
     int n;
     while(scanf("%d", &n)!=EOF && n)
     {
         for(int i = ; i < n; i++)
             scanf("%s", ch[i]);
         int ans = ;
         for(int i = ; i < n; i++)
         {
             for(int j = n-; j >= ; j--)
             {
                 if(i ==  || j == n-)
                 {
                     dp[i][j] = ;
                     continue;
                 }
                 int u, r, cnt = ;
                 for(int k = ; k <= dp[i-][j+]; k++){
                     u = i-k;
                     r = j+k;
                     if(ch[u][j] == ch[i][r])cnt++;
                     else break;
                 }
                 dp[i][j] = min(dp[i-][j+]+, cnt);
                 if(ans < dp[i][j])ans = dp[i][j];
             }
         }
         printf("%d\n", ans);
     }

     return ;
 }