codeforces 570 E. Pig and Palindromes (dp)

E. Pig and Palindromes

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Peppa the Pig was walking and walked into the forest. What a strange coincidence! The forest has the shape of a rectangle, consisting of nrows and m columns. We enumerate the rows of the rectangle from top to bottom with numbers from 1 to n, and the columns — from left to right with numbers from 1 to m. Let's denote the cell at the intersection of the r-th row and the c-th column as (r, c).

Initially the pig stands in cell (1, 1), and in the end she wants to be in cell (n, m). Since the pig is in a hurry to get home, she can go from cell (r, c), only to either cell (r + 1, c) or (r, c + 1). She cannot leave the forest.

The forest, where the pig is, is very unusual. Some cells of the forest similar to each other, and some look very different. Peppa enjoys taking pictures and at every step she takes a picture of the cell where she is now. The path through the forest is considered to be beautifulif photographs taken on her way, can be viewed in both forward and in reverse order, showing the same sequence of photos. More formally, the line formed by the cells in order of visiting should be a palindrome (you can read a formal definition of a palindrome in the previous problem).

Count the number of beautiful paths from cell (1, 1) to cell (n, m). Since this number can be very large, determine the remainder after dividing it by 109 + 7.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500) — the height and width of the field.

Each of the following n lines contains m lowercase English letters identifying the types of cells of the forest. Identical cells are represented by identical letters, different cells are represented by different letters.

Output

Print a single integer — the number of beautiful paths modulo 109 + 7.

Sample test(s)

input

3 4
aaab
baaa
abba

output

3

Note

Picture illustrating possibilities for the sample test.

比赛的时候想到了是dp搞...

不过dp废.....

可能更多的是心理上...

这道题并不怎么难想,但是以觉得是dp,就给了自己一种暗示...这题我搞不出来...

实际上,我把cA掉的时候应该还有一个小时十分钟左右的样子...

d题没啥思路,所以我有大概一个小时的时间可以搞e...未必就搞不出来.

还有因为答案很大要取模,感觉一般取模的题,要么是数学,要么是像dp这种有递推式子的.

这道题的思路是:

因为要形成回文串,我们可以从两边往中间走,保证每一步都相同.

dp[i][x1][x2] 表示路径长度为i,左上角出发到达x坐标为x1,又下角出发到达x坐标为x2,且两条路径上对应的字母都相同的方案数.

然后判断当前格子的字母是否一样,如果一样,则考虑转移.

由于从左上角出发可以往下往右,从右下角出发可以往上往左,排列组合,所以当前状态和之前的四种状态有关.

由因为这步的状态只和上以步的四种状态有关,所以路径长度那以维要滚动掉不然会MLE

dp方程为 dp[cur][x1][x2]=(dp[cur^1][x1][x2],dp[cur^1][x1-1][x2],dp[cur^1][x1][x2+1],dp[cur^1][x1-1][x2+1])%MOD;

然后注意由于(m+n) 的奇偶性,答案会有所不同.

根据奇偶性判断从两端出发是到两个相邻的格子还是到同一个格子.

初始化的话如果(1,1)和{n,m}点的字母一样那么 dp[0][1][n] 为1,否则为0.

其他点显然都为0

/*************************************************************************
    > File Name: code/cf/#316/EE.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月15日 星期六 04时10分13秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int MOD=1E9+7;
const int N=5E2+7;
int n,m;
char st[N][N];

int dp[2][N][N];
int main()
{
    scanf("%d %d",&n,&m);
    for ( int i = 1 ; i <= n ; i++)
    {
    scanf("%s",&st[i][1]);
    }
  //  dp[0][1][n]= st[1][1]==st[n][m];
    if (st[1][1]==st[n][m])
    {
    dp[0][1][n]=1;
    }
    else
    {
    dp[0][1][n]=0;
    }
    int cur = 0;
    int mx = (m+n-2)/2;
    for ( int step = 1 ; step <= mx ; step++)
    {
    cur = cur ^ 1;
    for (int i = 1 ; i<= n ; i++)
    {
        for ( int j = 1 ; j <= n ; j++ )
        {
        dp[cur][i][j]  =0 ;
        }
    }
    
     for ( int x1 = 1 ; x1 <= n&&x1-1<=step ;x1++)
     {
        for ( int x2 = n ; x2>=1 &&n-x2<=step ;x2--)
        {
            int y1 = 1+step-(x1-1);         
            int y2 = m-step+(n-x2);         //由x1,x2 可以计算出y1,y2
            if (st[x1][y1]==st[x2][y2])
            {
            dp[cur][x1][x2] = (dp[cur][x1][x2] + dp[cur^1][x1][x2])%MOD;
            dp[cur][x1][x2] = (dp[cur][x1][x2] + dp[cur^1][x1][x2+1])%MOD;
            dp[cur][x1][x2] = (dp[cur][x1][x2] + dp[cur^1][x1-1][x2])%MOD;
            dp[cur][x1][x2] = (dp[cur][x1][x2] + dp[cur^1][x1-1][x2+1])%MOD;
            }//只有当前pic 相同的时候才转移
        }   
     }
    }
     int ans = 0;
     for ( int i = 1 ; i<= n ; i++ )
     {
         ans = (ans + dp[cur][i][i]) % MOD;
     }
     if ((m+n)%2)
     {
        for ( int i = 1 ; i<= n-1 ; i++)
        {
        ans = (ans + dp[cur][i][i+1])%MOD;
        }
     }
     cout<<ans<<endl;
  
    return 0;
}