codeforces-149D（区间dp + 递归）

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最新推荐文章于 2024-02-02 23:32:57 发布

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分类专栏：区间dp

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本文探讨了给定正确括号序列的涂色问题，旨在找出所有可能的涂色方案，确保相邻涂色括号颜色不同，且每对括号至少有一方被涂色。通过动态规划方法，实现了高效的解决方案。

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题目：

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

Input

(()())

Output

Input

()

Output

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：给出一段匹配好的括号序列，给他们涂颜色，要么不涂色，要么只能涂红，蓝两种颜色，每两个相邻的括号不能涂相同的颜色，一个括号的左右必须有一个涂。求涂色方案数。

参考：https://blog.youkuaiyun.com/loy_184548/article/details/65448189

https://blog.youkuaiyun.com/u013534123/article/details/78095359

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>

using namespace std;

const int maxn = 705;
const int mod = 1e9 + 7;


int match[maxn];
long long dp[maxn][maxn][3][3];//dp[l][r][i][j] 表示l涂i，r涂j。
string str;

void getmatch()
{
    int n = str.size();
    stack <int> st;
    for(int i = 0;i < n;i ++)
    {
        if(str[i] == '(')
            st.push(i);
        else
        {
                match[i] = st.top();
                match[st.top()] = i;
                st.pop();
        }
    }
}

void dfs(int l,int r)
{
    if(l == r - 1)//一对括号
    {
        dp[l][r][0][1] = 1;
        dp[l][r][0][2] = 1;
        dp[l][r][1][0] = 1;
        dp[l][r][2][0] = 1;
        return ;
    }
    if(match[l] == r)//两个端点匹配
    {
        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(j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + 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(i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l + 1][r - 1][i][j]) % mod;
        }
    }
    else
    {
        int k = match[l];//从中间选取匹配的
        dfs(l,k);
        dfs(k + 1,r);
        for(int i = 0;i < 3;i ++)//l
            for(int j = 0;j < 3;j ++)//r
            for(int q = 0;q < 3;q ++)//k
            for(int p = 0;p < 3;p ++)//k + 1
        {
            if((q == 1 || q == 2) && (q == p))
                continue;
            dp[l][r][i][j] = (dp[l][r][i][j] + (dp[l][k][i][q] * dp[k + 1][r][p][j]) % mod) % mod;
        }
    }
}

int main()
{
    while(cin >> str)
    {
        int n = str.size();
        memset(dp,0,sizeof(dp));
        getmatch();
        dfs(0,n - 1);
        long long ans = 0;
        for(int i = 0;i < 3;i ++)
            for(int j = 0;j < 3;j ++)
            ans = (ans + dp[0][n - 1][i][j]) % mod;
        cout << ans << endl;
    }
    return 0;
}