Coloring Brackets
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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:
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
区间DP
用dp[i][j][p][k] 表示i到j的区间,左端点是什么颜色,右端点是什么颜色。用DFS实现比较简单一点,用递推很烦
关于区间DP,可以参照这个博客
http://blog.csdn.net/dacc123/article/details/50885903
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <stack>
using namespace std;
const long long int mod=1e9+7;
char a[705];
long long int dp[705][705][3][3];
int s[705];
int m[705];
int top;
void dfs(int i,int j)
{
if(j-i==1)
{
dp[i][j][0][1]=1;
dp[i][j][0][2]=1;
dp[i][j][1][0]=1;
dp[i][j][2][0]=1;
return;
}
else if(m[i]==j)
{
dfs(i+1,j-1);
for(int p=0;p<3;p++)
{
for(int q=0;q<3;q++)
{
if(q!=1) dp[i][j][0][1]=(dp[i][j][0][1]+dp[i+1][j-1][p][q])%mod;
if(q!=2) dp[i][j][0][2]=(dp[i][j][0][2]+dp[i+1][j-1][p][q])%mod;
if(p!=1) dp[i][j][1][0]=(dp[i][j][1][0]+dp[i+1][j-1][p][q])%mod;
if(p!=2) dp[i][j][2][0]=(dp[i][j][2][0]+dp[i+1][j-1][p][q])%mod;
}
}
return;
}
else
{
int k=m[i];
dfs(i,k);
dfs(k+1,j);
for(int p=0;p<3;p++)
for(int q=0;q<3;q++)
for(int x=0;x<3;x++)
for(int y=0;y<3;y++)
if(!((y==1&&x==1)||(y==2&&x==2)))
dp[i][j][p][q]=(dp[i][j][p][q]+(dp[i][k][p][x]*dp[k+1][j][y][q])%mod)%mod;
return;
}
}
int main()
{
while(scanf("%s",a)!=EOF)
{
int len=strlen(a);
top=-1;
for(int i=0;i<len;i++)
{
if(a[i]=='(')
s[++top]=i;
else
{
m[s[top]]=i;
//m[i]=s[top];
top--;
}
}
memset(dp,0,sizeof(dp));
dfs(0,len-1);
long long int 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;
}