HDU 4618Palindrome Sub-Array(暴力枚举每一个正方形)

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 933    Accepted Submission(s): 439

Problem Description

　　A palindrome sequence is a sequence which is as same as its reversed order. For example, 1 2 3 2 1 is a palindrome sequence, but 1 2 3 2 2 is not. Given a 2-D array of N rows and M columns, your task is to find a maximum sub-array of P rows and P columns, of which each row and each column is a palindrome sequence.

 

Input

　　The first line of input contains only one integer, T, the number of test cases. Following T blocks, each block describe one test case.
　　There is two integers N, M (1<=N, M<=300) separated by one white space in the first line of each block, representing the size of the 2-D array.
　　Then N lines follow, each line contains M integers separated by white spaces, representing the elements of the 2-D array. All the elements in the 2-D array will be larger than 0 and no more than 31415926.

 

Output

　　For each test case, output P only, the size of the maximum sub-array that you need to find.

 

Sample Input

   
   
   
   

    
    
    
    1
5 10
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 9 10 4 5 6 7 8
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
   
   
   
   

 

Source

2013 Multi-University Training Contest 2

 

                  题目大意： 给你一个矩形，让你找出最大的正方形边长。使得任意行回文任意列回文。不敢直接暴力做。当时就在想用manacher解决。但是manacher需要的是字符,这个是int。看了题解居然真的是暴力枚举，泪流满面。。

          解题思路：直接从边长最大开始从最小找。这点跟暴力KMP那个题目很像，然后就是暴力枚举每一个正方形的最左上的顶点，如果不满足，便移动这个顶点。详见代码。

          题目地址：Palindrome Sub-Array

AC代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int map1[305][305],n,m;

int solve(int len)
{
    int r1,r2,c1,c2,i,r11,r22,c11,c22;  //c表示行r表示列
    int flag1;
    for(c1=0;c1<=n-len;c1++)  //行顶点
    {
        for(r1=0;r1<=m-len;r1++)  //列顶点
        {
            flag1=0;
            r2=r1+len-1;
            c2=c1+len-1;
            for(i=c1;i<c1+len;i++)  //c1~c1+len-1行都需要满足
            {
               r11=r1,r22=r2;
               while(r11<=r22)
               {
                 if(map1[i][r11]!=map1[i][r22])
                 {
                    flag1=1;   //有一行不满足
                    break;
                 }
                 if(flag1) break;
                 r11++,r22--;
               }
               if(flag1) break;
            }
            if(flag1) continue;

            for(i=r1;i<r1+len;i++)  //每一行满足了，再看每一列
            {
                c11=c1,c22=c2;
                while(c11<=c22)
                {
                    if(map1[c11][i]!=map1[c22][i])
                    {
                        flag1=1;
                        break;
                    }
                    if(flag1) break;
                    c11++,c22--;
                }
                if(flag1) break;
            }
            if(flag1) continue;
            if(!flag1) return 1;
        }
    }
    return 0;
}

int main()
{
    int tes,i,j,mi;
    scanf("%d",&tes);
    while(tes--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                scanf("%d",&map1[i][j]);
        mi=n;  //mi记录min(n,m)即为可能的最大值
        if(mi>m) mi=m;
        for(i=mi;i>=1;i--)
        {
            if(solve(i))
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

//62MS 588K