HDU 4618 Palindrome Sub-Array

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 751 Accepted Submission(s): 366

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

Source

2013 Multi-University Training Contest 2

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zhuyuanchen520

刚开始想多了，今天看看，暴力模拟就可以了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>

using namespace std;

int n,m,num[320][320];

int ok(int x){
    int flag;
    for(int i=1;i<=n-x+1;i++){
        for(int j=1;j<=m-x+1;j++){
            flag=1;
            for(int l=i;l<=x+i-1 && flag;l++)
                for(int r=j;r<j+x/2 && flag;r++)
                    if(num[l][r]!=num[l][x+j-1-(r-j)])
                        flag=0;
            //printf("i=%d  j=%d   x=%d
",i,j,x);
            /*
            if(flag){
                printf("i=%d  j=%d                        x=%d
",i,j,x);
                printf("--------------------
");
                for(int l=i;l<x+i;l++){
                    for(int r=j;r<x+j;r++)
                        printf("%d ",num[l][r]);
                    printf("
");
                }
                printf("--------------------
");
            }
            */
            if(flag)
                return x;
        }
    }
    //printf("222
");
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&num[i][j]);
        int i;
        for(i=min(n,m);i>1;i--)
            if(ok(i)){
                printf("%d
",i);
                break;
            }
        if(i==1)
            printf("1
");
    }
    return 0;
}