Palindromic Paths（DP）

描述

Given an N×N grid of fields (1≤N≤500), each labeled with a letter in the alphabet. For example:

ABCD

BXZX
CDXB
WCBA

Each day, Susa walks from the upper-left field to the lower-right field, each step taking her either one field to the right or one field downward. Susa keeps track of the string that she generates during this process, built from the letters she walks across. She gets very disoriented, however, if this string is a palindrome (reading the same forward as backward), since she gets confused about which direction she had walked.

Please help Susa determine the number of distinct routes she can take that correspond to palindromes. Different ways of obtaining the same palindrome count multiple times. Please print your answer modulo 1,000,000,007.

输入

The first line of input contains N, and the remaining N lines contain the N rows of the grid of fields. Each row contains N characters that are in the range A...Z.

输出

Please output the number of distinct palindromic routes Susa can take, modulo 1,000,000,007.

样例输入

4
ABCD
BXZX
CDXB
WCBA

样例输出

 12

提示

Susa can make the following palindromes

1 x "ABCDCBA"

1 x "ABCWCBA"

6 x "ABXZXBA"

4 x "ABXDXBA"

 题目大意：

从左上角走到右下角（每次只能往右或往下）路径组成的串是回文串的路径数。

分析：可以让左下角和右上角同时开始走，dp[i][j][k]代表走i步左上角走到第j行，右上角走到第k行的回文数。dp[i][j][k]=dp[i-1][j-1][k]+dp[i-1][j-1][k+1]+dp[i-1][j][k]+dp[i-1][j][k+1]，由于n最大为500，三维开不下，观察状态转移方程第i步只于第i-1步有关系，所以可以用滚动数组来优化。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int MD=1e9+7;
char s[505][505];
ll dp[2][505][505];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]+1);
    int cnt=0;
    dp[cnt][1][n]=1;
    for(int i=1;i<=n;i++)
    {
        for(int x1=1;x1<=n;x1++)
            for(int x2=1;x2<=n;x2++)
            {
                int y1=i-x1+1,y2=2*n-x2-i+1;
                if(y1<=0||y1>n||y2<=0||y2>n) continue;
                if(s[x1][y1]==s[x2][y2])
                    dp[cnt^1][x1][x2]=(dp[cnt][x1-1][x2]+dp[cnt][x1-1][x2+1]+dp[cnt][x1][x2]+dp[cnt][x1][x2+1])%MD;
            }
        for(int i=1;i<=n;i++)///很重要
            for(int j=1;j<=n;j++)
                dp[cnt][i][j]=0;
        cnt^=1;
    }
    ll ans=0;
    for(int i=1;i<=n;i++)
        ans+=dp[cnt][i][i];
    printf("%I64d
",ans%MD);
    return 0;
}