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 = 1010;
char ch[N][N];
int dp[N][N];

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF && n)
    {
        for(int i = 0; i < n; i++)
            scanf("%s", ch[i]);
        int ans = 1;
        for(int i = 0; i < n; i++)
        {
            for(int j = n-1; j >= 0; j--)
            {
                if(i == 0 || j == n-1)
                {
                    dp[i][j] = 1;
                    continue;
                }
                int u, r, cnt = 1;
                for(int k = 1; k <= dp[i-1][j+1]; k++){
                    u = i-k;
                    r = j+k;
                    if(ch[u][j] == ch[i][r])cnt++;
                    else break;
                }
                dp[i][j] = min(dp[i-1][j+1]+1, cnt);
                if(ans < dp[i][j])ans = dp[i][j];
            }
        }
        printf("%d
", ans);
    }

    return 0;
}