Coloring Brackets CodeForces - 149D（区间dp+dfs）

Coloring Brackets CodeForces - 149D

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).

Examples
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：

　　给一个合法的括号串，然后问这串括号有多少种涂色方案，涂色是有限制的。

　　1,每个括号只有三种选择：涂红色，涂蓝色，不涂色。

　　2,每对括号有且仅有其中一个被涂色。

　　3,相邻的括号不能涂相同的颜色，但是相邻的括号可以同时不涂色。

分析：

给出的括号序列，括号的匹配都是固定的，也就是说只需要给这些固定匹配的括号按照上面限制涂色就行了。

定义 dp[l][r][x][y] d p [ l ] [ r ] [ x ] [ y ] 表示区间 [l,r] [ l , r ] 在左端点涂x色，右端点涂y色的情况下的方案数。其中0代表不涂色， 1代表涂红色， 2代表涂蓝色。

可以把括号序列可以分为两类分别进行状态转移：

括号套括号，状态转移是：
dp[l][r][x][y]+=dp[l+1][r−1][x′][y′](0<=x′<3,x′!=x,0<=y′<3,y!=y′) d p [ l ] [ r ] [ x ] [ y ] + = d p [ l + 1 ] [ r − 1 ] [ x ′ ] [ y ′ ] ( 0 <= x ′ < 3 , x ′ ! = x , 0 <= y ′ < 3 , y ! = y ′ )

（即最外面两个括号匹配）

多个匹配串连起来，转台转移是：
dp[l][r][x][y]+=dp[l][nu][x′][y′]∗dp[nu+1][r][x′′][y′′] d p [ l ] [ r ] [ x ] [ y ] + = d p [ l ] [ n u ] [ x ′ ] [ y ′ ] ∗ d p [ n u + 1 ] [ r ] [ x ″ ] [ y ″ ] (nu是l对应的另一边括号)

（即lr最外侧两个括号不匹配，需要通过深搜将他们分成一个一个匹配的小段，然后在加起来）

最终

当l+1 == r的时候有： dp[l][r][0][1]=1;　　dp[l][r][1][0]=1; d p [ l ] [ r ] [ 0 ] [ 1 ] = 1 ; 　 　 d p [ l ] [ r ] [ 1 ] [ 0 ] = 1 ;

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

code：

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
const ll maxn = 710;
ll dp[maxn][maxn][3][3],match[maxn],tep[maxn];
void get_match(char a[]){//模拟栈进行括号匹配
    int p = 0;
    for(int i = 0; a[i]; i++){
        if(a[i] == '(')//如果是左括号入栈
            tep[p++] = i;
        else{//如果是右括号，那么和栈顶对应，分别记录匹配位置
            match[i] = tep[p-1];
            match[tep[p-1]] = i;
            p--;//出栈
        }
    }
}
void dfs(int l,int r){
    if(l + 1 == r){//如果枚举到了最里面的匹配括号，只有以下四种情况
        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 ;
    }
    else if(match[l] == r){//如果当前的lr匹配,当前的最优值为原本数量加内部最优
        dfs(l+1,r-1);
        for(int i = 0;  i < 3; i++){
            for(int j = 0; j < 3; j++){
                if(j != 1)//相邻颜色不同
                    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 ;
    }
    else{//如果当前不匹配，则分成两部分，前一部分匹配的和后一部分，后一部分如果不匹配又可再分不断递归下去分成多段匹配的
        int nu = match[l];
        dfs(l,nu);
        dfs(nu+1,r);
        for(int i = 0; i < 3; i++){
            for(int j = 0; j < 3; j++){
                for(int x = 0; x < 3; x++){
                    for(int y = 0; y < 3; y++){
                        if(!(x == 1 && y == 1 || x == 2 && y == 2)){//相邻颜色不同
                            dp[l][r][i][j] = (dp[l][r][i][j] + dp[l][nu][i][x] * dp[nu+1][r][y][j] % mod) % mod;
                        }
                    }
                }
            }
        }
    }
}
int main(){
    char s[maxn];
    while(~scanf("%s",s)){
        memset(dp,0,sizeof(dp));
        int len = strlen(s);
        get_match(s);
        dfs(0,len-1);
        ll ans = 0;
        for(int i = 0; i < 3; i++){
            for(int j = 0; j < 3; j++){
                ans = (ans + dp[0][len-1][i][j]) % mod;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}