hdu4618(KMP+动态规划)

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 317    Accepted Submission(s): 166

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

15 101 2 3 3 2 4 5 6 7 81 2 3 3 2 4 5 6 7 81 2 3 3 2 4 5 6 7 81 2 3 3 2 4 5 6 7 81 2 3 9 10 4 5 6 7 8

 

Sample Output

4
 
 
本题在比赛的时候没有想到好的方法，索性没做，赛后相结合KMP算法运用动态规划思想，不了能力太弱，也没能动手，结果就写了一个5重循环，竟能过
后来索性提交了几次，得到了一些无关技巧的小技巧，裸函数、宏定义、内联函数对运行时间的影响
#include<iostream>#include<cstdio>using namespace std;//#define Min(a,b) a<b?a:b//#define Max(a,b) a>b?a:bconst int MAX=300+10;int da[MAX][MAX];//宏定义Max(a,b),Min(a,b)，运行时间为512Ms//直接写裸函数int Min(int a,int b),int Max(int a,int b)，运行时间为656Ms//写内联函数inline int Min(int a,int b)，inline int Max(int a,int b)，运行时间为525Msint Min(int a,int b){return a<b?a:b;}int Max(int a,int b){return a<b?b:a;}inline int Judge(int x,int y,int len){int i,j,k;for(k=y;k<=y+len-1;k++){for(i=x,j=x+len-1;i<=j;i++,j--){if(da[i][k]!=da[j][k])return 0;}}for(k=x;k<=x+len-1;k++){for(i=y,j=y+len-1;i<=j;i++,j--){if(da[k][i]!=da[k][j])return 0;}}return len;}int main(){int cas,i,j,n,m,ans,len,tmp;cin>>cas;while(cas--){scanf("%d%d",&n,&m);for(i=0;i<n;i++){for(j=0;j<m;j++)scanf("%d",&da[i][j]);}ans=0;for(i=0;i<n;i++){for(j=0;j<m;j++){len=Min(n-i,m-j)+1;do{tmp=Judge(i,j,len);ans=Max(ans,tmp);}while(--len);}}printf("%d\n",ans);}return 0;}