HDU 2859 Phalanx（对称矩阵 经典dp样例）

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=2859

Phalanx

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2792    Accepted Submission(s): 1357

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

 

题目意思：

给你一个字符矩阵，比如n*n的矩阵，问你最大的对称矩阵的阶数是多少

阶数：比如3*3的矩阵，阶数就是3

对称矩阵的定义：副对角线两边的字符对称，副对角线：从左下到右上

 

分析：

我们从大矩阵的右上角看，每次矩阵阶数加一，都是原来矩阵的下面加一行，左边加一列

给一个2阶的对称矩阵，它对称的字符肯定只有一对

给你一个三阶的对称矩阵，它的最下面的行和最左边的列，对称字符的个数肯定是两对

同理，给一个四阶的对称矩阵，它的最下面的行和最左边的列，对称字符的个数肯定是三对

所以如果该矩阵是对称矩阵的话，随着它阶数的加1，它最下面的行和最左的列的对称字符数也加一

这是后面能dp的原因

 

dp【i】【j】：以i，j为矩阵的左下角，最大对称矩阵的阶数

 

如果当前矩阵i，j的最下一行和最左一列的字符数大于矩阵i-1，j+1的最下一行和最左一列的字符数的话，那么当前矩阵就有可能的对称矩阵（有可能）

所以： dp[i][j]=dp[i-1][j+1]+1;

 

记住：我们是从右上角开始推的，往左下角移动

code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 1005
char a[max_v][max_v];
int dp[max_v][max_v];
int main()
{
    int n;
    while(cin>>n,n)
    {
        memset(dp,0,sizeof(dp));
        getchar();
        for(int i=0;i<n;i++)
        {
            scanf("%s",&a[i]);
            dp[0][i]=1;//初始化
        }
        int ans=1;
        for(int i=1;i<n;i++)
        {
            for(int j=n-1;j>=0;j--)//右上角开始
            {
                int x=i-1,y=j+1;//上一个矩阵的左下角坐标
                int num=1;//对称字符个数
                while(x>=0&&x<n&&y>=0&&y<n&&a[i][y]==a[x][j])//坐标在大矩阵内且字符对称
                {
                    x--;//横坐标上移
                    y++;//纵坐标右移
                    num++;//对称字符数加1
                }
                if(num>dp[i-1][j+1])//判断当前矩阵是不是对称矩阵
                {
                    dp[i][j]=dp[i-1][j+1]+1;
                }else
                {
                    dp[i][j]=num;
                }
                 ans=max(ans,dp[i][j]);//找对称矩阵最大阶
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}