HDU 2859 Phalanx （DP）

Phalanx

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 363    Accepted Submission(s): 170

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

 

Recommend

gaojie

 

对于每个字符看该列以上和该行右侧的字符匹配量，如果匹配量大于右上角记录下来的矩阵大小，就是右上角的数值+1，否则就是这个匹配量。

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

int dp[1010][1010];
char str[1010][1010];

int main()
{
    int n;
    while(scanf("%d",&n) == 1 && n)
    {
        for(int i = 0;i < n;i++)
            scanf("%s",str[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 t1 = i, t2 = j;
                while(t1 >= 0 && t2 < n && str[t1][j] == str[i][t2])
                {
                    t1--;
                    t2++;
                }
                t1 = i - t1;
                if(t1 >= dp[i-1][j+1]+1)dp[i][j] = dp[i-1][j+1]+1;
                else dp[i][j] = t1;
                ans = max(ans,dp[i][j]);
            }
        printf("%d
",ans);
    }
    return 0;
}