HDU 2859 Phalanx （DP）

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

状态：dp[i][j]为第i行第j列所能够组成的最大对称子矩阵的长度。关于对角线完全对称的矩阵！

转移方程为：dp[i][j]=dp[i-1][j+1]+1 ； 注意这里是由点（i-1，j+1）推过来的。因为我们在这里矩阵的对角线是由左下角推到右上角去的。

首先我们要进行初始化，第0行的它们所能组成的dp[0][i]=1，最大只能组成1个。

然后我们对每个点进行判断(i,j)，当然它是从(i-1,j+1)推过来的。但是对于每个字符看该列以上和该行右侧的字符匹配量，如果该匹配量大于右上角记录下来的矩阵大小，那么就是右上角的数值加1，否则就是这个匹配量（因为我们我们每次都要满足所有的，所以要尽可能的取小的）

参考网址：here

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define maxn 1111
int dp[maxn][maxn];
char arr[maxn][maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        int ans=1;
        for(int i=0; i<n; i++) scanf("%s",arr[i]);
        for(int i=0; i<n; i++) dp[0][i]=1;
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                int tmpx=i,tmpy=j;
                while( tmpx>=0 && tmpy<n && arr[tmpx][j]==arr[i][tmpy] )
                    tmpx--, tmpy++;
                dp[i][j] = min(dp[i-1][j+1]+1,i-tmpx);
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}

稍微改动下， 注意边界细节！！！改完之后感觉没有上面代码清爽！！！

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define maxn 1111
int dp[maxn][maxn];
char arr[maxn][maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        int ans=1;
        for(int i=0; i<n; i++) scanf("%s",arr[i]);
        for(int i=0; i<n; i++) dp[0][i]=1;
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                int tmpx=i,tmpy=j;
                while( tmpx>=0 && tmpy<n && arr[tmpx][j]==arr[i][tmpy] && dp[i-1][j+1]+1>i-tmpx)///
                    tmpx--, tmpy++;
                dp[i][j] = i-tmpx; ///tmpx最后所指向的不符合条件的位置，这样就不用加1了（其实长度应该是i-符合条件的位置）
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}