51Nod 1503 猪和回文

#include <stdio.h>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#define Faster ios::sync_with_stdio(false),cin.tie(0)
#define Read freopen("in.txt", "r", stdin),freopen("out.txt", "w", stdout)
const int INF = 0x3f3f3f3f;
const int maxn = 500 + 5;
const int MOD = 1e9 + 7;

char g[maxn][maxn];

//f[a][b][c][d]        :从(1,1)->(a,b)和从(n,m)->(c,d);
//f[a][b][c]        :知道a,b,c就能得出d所以降了一个维
//f[s][][]  由f[s-1][][]得出，所以可以用滚动数组
ll dp[2][maxn][maxn];

//检查是否满足回文的条件
bool check(int a, int b, int c, int d){
    if(a == c && b == d) return true;
    if(a+1 == c && b == d) return true;
    if(a == c && b+1 == d) return true;
    return false;
}

int main() 
{
    Faster;
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1;i <= n;i++){
        scanf("%s", g[i]+1);
    }
    if(g[1][1] == g[n][m]){
        dp[1][1][n] = 1;
    }
    ll ans = 0;
    for(int i = 1;i <= n;i++){
        for(int j = 1; (i+j-1) <= (n+m)/2;j++){
            for(int k = n;n+m+2-i-j-k <= m;k--){
                int z = n+m+2-i-j-k;
                if(g[i][j] == g[k][z]){
                    dp[i&1][j][k] += dp[(i-1)&1][j][k+1];
                    dp[i&1][j][k] += dp[(i-1)&1][j][k];
                    dp[i&1][j][k] += dp[i&1][j-1][k+1];
                    dp[i&1][j][k] += dp[i&1][j-1][k];
                    dp[i&1][j][k] %= MOD;
                    if(check(i,j,k,z)){
                        ans = (ans + dp[i&1][j][k])%MOD;
                    }
                }
            }
        }
        memset(dp[(i-1)&1], 0, sizeof(dp[(i-1)&1]));
    }
    cout << ans << endl;
    return 0;
}