HDU2859 Phalanx DP

Phalanx

Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

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

题解

一开始想了很久没想出来…脑子还是不够灵活，搜了很多题解才慢慢明白做法。

对每个点对它上面和右边的点进行匹配，如果匹配量大于右上角记录的矩阵最大的值，那么这个值就+1，否则就等于匹配量

代码

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

char mar[1001][1001];
int dp[1001][1001];
int main()
{
int n;
while (scanf("%d",&n)&& n)
{
for (int i = 0;i<n;i++)
scanf("%s",&mar[i]   );
int maxn = 1;
for (int i = 0;i<n;i++)
{
for (int j = 0;j<n;j++)
{
if (i == 0 || j == n-1)
{
dp[i][j] = 1;
continue;
}

int t1 = i;int t2 = j;
while (t1>=0 && t2<n && mar[t1][j] == mar[i][t2])
{
t1--;
t2++;
}
int a = i-t1;
if (a >= dp[i-1][j+1]+1)
dp[i][j] = dp[i-1][j+1]+1;
else
dp[i][j] = a;
if (dp[i][j] > maxn)
maxn = dp[i][j];
}
}
printf("%d\n",maxn);
memset(dp,0,sizeof(dp));
memset(mar,0,sizeof(mar));
}
}