CF 149D 区间dp

Once Petya read a problem about a bracket sequence. He gave it much thought but didn't find a solution. Today you will face it.

You are given string s. It represents a correct bracket sequence. A correct bracket sequence is the sequence of opening ("(") and closing (")") brackets, such that it is possible to obtain a correct mathematical expression from it, inserting numbers and operators between the brackets. For example, such sequences as "(())()" and "()" are correct bracket sequences and such sequences as ")()" and "(()" are not.

In a correct bracket sequence each bracket corresponds to the matching bracket (an opening bracket corresponds to the matching closing bracket and vice versa). For example, in a bracket sequence shown of the figure below, the third bracket corresponds to the matching sixth one and the fifth bracket corresponds to the fourth one.

You are allowed to color some brackets in the bracket sequence so as all three conditions are fulfilled:

• Each bracket is either not colored any color, or is colored red, or is colored blue.
• For any pair of matching brackets exactly one of them is colored. In other words, for any bracket the following is true: either it or the matching bracket that corresponds to it is colored.
• No two neighboring colored brackets have the same color.

Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. Two ways of coloring are considered different if they differ in the color of at least one bracket. As the result can be quite large, print it modulo 1000000007 (10⁹ + 7).

Input

The first line contains the single string s (2 ≤ |s| ≤ 700) which represents a correct bracket sequence.

Output

Print the only number — the number of ways to color the bracket sequence that meet the above given conditions modulo 1000000007 (10⁹ + 7).

Example

Input

(())

Output

12

Input

(()())

Output

40

Input

()

Output

4

Note

Let's consider the first sample test. The bracket sequence from the sample can be colored, for example, as is shown on two figures below.

 

The two ways of coloring shown below are incorrect.

 

题意大概就是

给一个给定括号序列，给该括号上色，上色有三个要求

1、只有三种上色方案，不上色，上红色，上蓝色

2、每对括号必须且只能给其中的一个上色

3、相邻的两个不能上同色，可以都不上色

求0-len-1这一区间内有多少种上色方案

区间dp果然还是改加维数加维数，因为两维根本不能控制好状态，因为收到颜色的影响的。

所以用4维表示。。。

dp[l][r][i][j]表示l-r区间两端颜色分别是i,j的方案数

0代表不上色，1代表上红色，2代表上蓝色

这样就可以很好的描述状态了。可以写出状态转移方程来了。

对于l-r区间，有3种情况

1、if(l+1==r) 说明就只有一对，那么1对括号涂色方法很简单，4种：

dp[l][r][0][1]=1;
        dp[l][r][1][0]=1;
        dp[l][r][0][2]=1;
        dp[l][r][2][0]=1;

2、if(l与r是配对的)

递归（l+1，r-1）得到dp[l+1][r-1][i][j]的值。

状态转移：4种，注意转移的时候相邻的括号颜色不能相同。

1.dp[l][r][0][1]=(dp[l][r][0][1]+dp[l+1][r-1][i][j])%mod; 

2.dp[l][r][1][0]=(dp[l][r][1][0]+dp[l+1][r-1][i][j])%mod;

3.dp[l][r][0][2]=(dp[l][r][0][2]+dp[l+1][r-1][i][j])%mod; 

4.dp[l][r][2][0]=(dp[l][r][2][0]+dp[l+1][r-1][i][j])%mod;

3、if(l与r不配对)

k是与i匹配的括号的位置。

注意转移的时候同样相邻的括号颜色不能相同。

dp[l][r][i][j]=(dp[l][r][i][j]+(dp[l][k][i][i1]*dp[k+1][r][j1][j])%mod)%mod;

发现dp写法其实都能递归来写，怎么样好写怎么样来吧，这种应该是。

#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <time.h>using namespace std;typedef long long LL;const int inf = 1e9;const int mod = 1e9+7;const int MAXN = 700+7;int n,m;char s[MAXN];LL dp[MAXN][MAXN][3][3];int match[MAXN];int sta[MAXN];void dfs(int l,int r){    if(r - l == 1)    {        dp[l][r][0][1] = 1;        dp[l][r][1][0] = 1;        dp[l][r][0][2] = 1;        dp[l][r][2][0] = 1;        return ;    }    if(match[l] == r)    {        dfs(l+1,r-1);        for(int i = 0 ; i < 3; ++i)            for(int j = 0 ; j < 3; ++j)        {            if(j != 1)dp[l][r][0][1] = (dp[l][r][0][1] + dp[l+1][r-1][i][j])%mod;            if(i != 1)dp[l][r][1][0] = (dp[l][r][1][0] + dp[l+1][r-1][i][j])%mod;            if(j != 2)dp[l][r][0][2] = (dp[l][r][0][2] + dp[l+1][r-1][i][j])%mod;            if(i != 2)dp[l][r][2][0] = (dp[l][r][2][0] + dp[l+1][r-1][i][j])%mod;        }        return ;    }    int k = match[l];    dfs(l,k);    dfs(k+1,r);    for(int i = 0 ; i < 3; ++i)        for(int j = 0 ; j < 3 ; ++j)            for(int i1 = 0 ; i1 < 3 ; ++i1)                for(int j1 = 0 ; j1 < 3; ++j1)                {                    if(i1 == j1 && i1 !=0)continue;                    dp[l][r][i][j] = (dp[l][r][i][j] + (dp[l][k][i][i1]*dp[k+1][r][j1][j])%mod)%mod;                }}int main(){    scanf("%s",s);    int l = strlen(s);    int top = 0;    for(int i = 0 ; i < l ; ++i)if(s[i] == '(')sta[top++] = i;    else match[sta[--top]] = i;    dfs(0,l-1);    LL ans = 0;    for(int i = 0 ; i < 3; ++i)        for(int j = 0 ; j < 3 ; ++j)    {        ans = (ans + dp[0][l-1][i][j])%mod;    }    printf("%I64d\n",ans);    return 0;}