hdu 2859 Phalanx (最大对称子矩阵)

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

 题意：找到最大对称的矩阵

每次只需求最外面一层对称个数sum，再和右上角对称矩阵大小加一取最小就行，就求出当前小矩阵的最大对称矩阵。最后取个所有对称矩阵大小的最大值就行。

dp[i][j] = min(sum,dp[i-1][j+1]+1);

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int n;
char G[1007][1007];
int dp[1007][1007];
int main(){
    ios::sync_with_stdio(false);
    while(cin>>n&&n){
        char a;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                cin>>a;
                if(a>='A'&&a<='Z')
                    a+=32;
                G[i][j]=a;
            }
        memset(dp,0,sizeof(dp));
        int ans=-inf;
        for(int i=1;i<=n;i++)
            for(int j=n;j>=1;j--){
                int x=i; int y=j;
                int sum=0;
                while(x>=1&&y<=n&&G[x][j]==G[i][y]){
                    x--; y++;
                    sum++;
                }
                dp[i][j]=min(sum,dp[i-1][j+1]+1);
                ans=max(ans,dp[i][j]);
            }
        cout<<ans<<endl;
    }
}