HDU-2859 Phalanx ( DP )

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2859

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

 

题目大意 给一个方阵，找一个最大子阵，该子阵满足关于从左下到右上的线对称，参照样例

思路 按问题描述可知，一个满足条件的子阵，其延对角线的子阵也是满足条件的，若以dp[i][j]表示map[i][j]为左下点的最大满足条件的方阵，则只需要比较第i行和第j列就能得到dp[i][j]和dp[i - 1][j + 1]的关系了，思路很简单，需要注意边界情况的判断和初始化，还有ans最好初始化为1

 

#include<iostream>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;

char map[1005][1005];
int dp[1005][1005];

int main(){
    ios::sync_with_stdio( false );

    int n;
    while( cin >> n, n ){
        for( int i = 0; i < n; i++ )
            cin >> map[i];
        int ans = 1;

        for( int i = 0; i < n; i++ ){
            for( int j = 0; j < n; j++ ){
                if( i == 0 || j == n - 1 ){
                    dp[i][j] = 1;
                    continue;
                }

                int x = i;
                int y = j;

                while( x >= 0 && y < n && map[x][j] == map[i][y] ){
                    x--;
                    y++;
                }

                if( i - x > dp[i - 1][j + 1] )
                    dp[i][j] = dp[i - 1][j + 1] + 1;
                else dp[i][j] = i - x;

                ans = max( ans, dp[i][j] );
            }
        }

        cout << ans << endl;
    }

    return 0;
}