HDU2859 Phalanx （DP）

不念不忘少年蓝@ 2022-07-21 08:44 84阅读 0赞

# Phalanx #

**Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 1093 Accepted Submission(s): 522**

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 ][2009 Multi-University Training Contest 5 - Host by NUDT]

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#include <stdio.h>
    #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\n",ans);
        }
        return 0;
    }

[2009 Multi-University Training Contest 5 - Host by NUDT]: http://acm.hdu.edu.cn/search.php?field=problem&key=2009+Multi-University+Training+Contest+5+-+Host+by+NUDT&source=1&searchmode=source