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 modulo1000000007 (109 + 7).
The first line contains the single string s (2 ≤ |s| ≤ 700) which represents a correct bracket sequence.
Print the only number — the number of ways to color the bracket sequence that meet the above given conditions modulo 1000000007(109 + 7).
(())
12
(()())
40
()
4
題目大意：
給一個給定合法的括號序列，給該序列每個括號上色，上色有三個要求
1、只有三種上色方案，不上色，上紅色，上藍色
2、每對匹配的括號必須只能給其中的一個上色
3、相鄰的兩個不能上同色，可以都不上色
分析：
區間dp，假設不上色爲0，上紅色爲1，上藍色爲2，我們可以設置狀態：
dp[cl][cr][l][r]:在l-1處染cl顏色，在r+1處染cr顏色，那麼區間[l,r]的染色情況數。
既然是區間dp，就應該找能夠分割區間的依據，那麼就是對於區間[l,r],找到位置l處的'('所以對應的')'的位置match[l],設爲R，那麼就可以遞歸求解[l+1,R-1]和[R+1,r]，然後遞歸的時候分別對兩個區間的兩邊的顏色cl、cr討論一下就好了，因爲題目的限制條件多，所以這裏的情況反而不多了。
用棧預處理出match數組即可。
/****************
*PID:149d div2
*Auth:Jonariguez
*****************
dp[cl][cr][l][r]:在l-1處染cl顏色，在r+1處染cr顏色，那麼區間[l,r]的染色情況數
cl={0,1,2}
答案爲dp[0][0][1][n]
記憶化搜索
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}
const int maxn=700+10;
const int MOD=1e9+7;
LL dp[3][3][maxn][maxn];
char str[maxn];
int match[maxn];
LL DP(int cl,int cr,int l,int r){
if(l==r+1 && cl==cr && cl!=0) return 0;
if(l>=r) return 1;
if(dp[cl][cr][l][r]>=0) return dp[cl][cr][l][r];
LL &ans=dp[cl][cr][l][r];
ans=0;
int R=match[l];
if(cl){
ans=(ans+DP(3-cl,0,l+1,R-1)*DP(0,cr,R+1,r)%MOD)%MOD;
ans=(ans+DP(0,1,l+1,R-1)*DP(1,cr,R+1,r)%MOD)%MOD;
ans=(ans+DP(0,2,l+1,R-1)*DP(2,cr,R+1,r)%MOD)%MOD;
}else {
ans=(ans+DP(1,0,l+1,R-1)*DP(0,cr,R+1,r)%MOD)%MOD;
ans=(ans+DP(2,0,l+1,R-1)*DP(0,cr,R+1,r)%MOD)%MOD;
ans=(ans+DP(0,1,l+1,R-1)*DP(1,cr,R+1,r)%MOD)%MOD;
ans=(ans+DP(0,2,l+1,R-1)*DP(2,cr,R+1,r)%MOD)%MOD;
}
// printf("cl=%d cr=%d l=%d r=%d ans=%d\n",cl,cr,l,r,ans);
return ans;
}
int main()
{
int i,j,n;
while(scanf("%s",str+1)!=EOF){
n=strlen(str+1);
for(i=0;i<=n;i++) match[i]=n+2; //即無窮大
stack<int> s;
for(i=1;i<=n;i++){
if(str[i]=='(')
s.push(i);
else {
match[s.top()]=i;
s.pop();
}
}
memset(dp,-1,sizeof(dp));
printf("%I64d\n",DP(0,0,1,n));
}
return 0;
}