Codeforces-149D-Coloring Brackets【区间DP】

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).

input

(())

output

12

input

(()())

output

40

input

()

output

4

题目链接：cf-149D

题目大意：

给一个给定括号序列，给该括号上色。

上色规则：
1.只能选择红色、蓝色、无色
2.每对括号，只能给其中一个上色
3.相邻两个颜色不可相同，但可以都为无色

题目思路：区间dp

0：无色； 1：红色； 2：蓝色

dp[i][j][p][q] //表示i-j这个区间，i涂p颜色，j涂q颜色的方案数

初始化：如果 l = r - 1;则表示只有一对括号

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

如果 l 和 r 是一对括号

dp[l][r][0][1] += dp[l + 1][r - 1][i][j]
dp[l][r][0][2] += dp[l + 1][r - 1][i][j]
dp[l][r][1][0] += dp[l + 1][r - 1][i][j]
dp[l][r][2][0] += dp[l + 1][r - 1][i][j]

如果不是，

dp[l][r][i][j] += dp[l][k][i][q] * dp[k + 1][r][p][j]
//k:和l配对的括号的位置；q:k所选的颜色；p：k+1所选的颜色

以下是代码：

#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <deque>
#include <list>

using namespace std;
#define mod 1000000007
#define maxn 705
long long dp[maxn][maxn][3][3];  //i到j区间，两端分别染成q,p颜色的方法数。q（0~3）
int match[maxn];

string s;
void getmatch()
{
int len = s.size();
stack<int> st;
for (int i = 0; i < len; i++)
{
if (s[i] == '(') st.push(i);
else
{
match[i] = st.top();
match[st.top()] = i;
st.pop();
}
}
}
void solve(int l, int r)
{
if (l == r - 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)
{
solve(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];  //找到与l匹配的点,记为k

solve(l,k);  //左边
solve(k + 1, r);  //右边

for (int i = 0; i < 3; i++)  //l的颜色
{
for (int j = 0; j < 3; j++)  //r的颜色
{
for (int q = 0; q < 3; q++) //k的颜色
{
for (int p = 0; p < 3; p++)  //k+1的颜色
{
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 >> s)
{
memset(dp, 0, sizeof dp);

int len = s.size();
getmatch();
solve(0,len - 1);
long long 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;

cout << ans << endl;

}
return 0;
}