[bzoj4098] [Usaco2015 Open]Palindromic Paths

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define ui unsigned int
 7 using namespace std;
 8 const int modd=1e9+7;
 9 ui f[2][505][505];
10 char mp[505][505],s[505][505];
11 int i,j,k,n,m;
12 int ans;
13  
14 int ra;char rx;
15 inline int read(){
16     rx=getchar(),ra=0;
17     while(rx<‘0‘||rx>‘9‘)rx=getchar();
18     while(rx>=‘0‘&&rx<=‘9‘)ra*=10,ra+=rx-48,rx=getchar();return ra;
19 }
20  
21 int main(){
22     n=read();
23     for(i=1;i<=n;i++)scanf("%s",s[i]+1);
24     if(s[1][1]!=s[n][n]){puts("0");return 0;}
25     for(i=1;i<=n+n-1;i++)
26         for(j=1;j<=min(i,n+n-i);j++)
27             if(i<=n)mp[i][j]=s[i-j+1][j];
28             else mp[i][j]=s[n-j+1][i+j-n];
29     bool now=1,pre=0;int nn=n+n;
30     f[0][1][1]=s[1][1]==s[n][n];
31     for(i=2;i<=n;i++,swap(now,pre))
32         for(j=1;j<=i;j++)for(k=1;k<=i;k++)if(mp[i][j]==mp[nn-i][k])
33             f[now][j][k]=(f[pre][j-1][k-1]+f[pre][j-1][k]+f[pre][j][k-1]+f[pre][j][k])%modd;
34         else f[now][j][k]=0;
35     for(i=1;i<=n;i++)ans+=f[pre][i][i],ans-=ans>=modd?modd:0;
36     printf("%d\n",ans);
37 }