CF149D Coloring Brackets

区间dp的题.

思路

可先用stake将每个位置的括号的匹配找出来,我们把他叫love数组
即

lovei=jlovej=i l o v e i = j l o v e j = i

(当然他们匹配)
我们可以开一个数组叫dp[i][j][0/1/2][0/1/2]
表示在左端点为i右端点为j左端点染了0/1/2右端点染了0/1/2时的方案数
分类讨论

• 当二者相邻,
  dp[l][r][0][1]=dp[l][r][0][2]=dp[l][r][1][0]=dp[l][r][2][0]=1; d p [ l ] [ r ] [ 0 ] [ 1 ] = d p [ l ] [ r ] [ 0 ] [ 2 ] = d p [ l ] [ r ] [ 1 ] [ 0 ] = d p [ l ] [ r ] [ 2 ] [ 0 ] = 1 ;
• 否则,当二者匹配

for i=0,i−>2,i++for j=0,j−>2;j++if(i!=1)dp[l][r][1][0]=(dp[l][r][1][0]+dp[l+1][r−1][i][j]);if(j!=1)dp[l][r][0][1]=(dp[l][r][0][1]+dp[l+1][r−1][i][j]);if(j!=2)dp[l][r][0][2]=(dp[l][r][0][2]+dp[l+1][r−1][i][j]);if(i!=2)dp[l][r][2][0]=(dp[l][r][2][0]+dp[l+1][r−1][i][j]); f o r   i = 0 , i − > 2 , i + + f o r   j = 0 , j − > 2 ; j + + i f ( i ! = 1 ) d p [ l ] [ r ] [ 1 ] [ 0 ] = ( d p [ l ] [ r ] [ 1 ] [ 0 ] + d p [ l + 1 ] [ r − 1 ] [ i ] [ j ] ) ; i f ( j ! = 1 ) d p [ l ] [ r ] [ 0 ] [ 1 ] = ( d p [ l ] [ r ] [ 0 ] [ 1 ] + d p [ l + 1 ] [ r − 1 ] [ i ] [ j ] ) ; i f ( j ! = 2 ) d p [ l ] [ r ] [ 0 ] [ 2 ] = ( d p [ l ] [ r ] [ 0 ] [ 2 ] + d p [ l + 1 ] [ r − 1 ] [ i ] [ j ] ) ; i f ( i ! = 2 ) d p [ l ] [ r ] [ 2 ] [ 0 ] = ( d p [ l ] [ r ] [ 2 ] [ 0 ] + d p [ l + 1 ] [ r − 1 ] [ i ] [ j ] ) ;

• 二者不匹配
  

  for(LL i=0;i<=2;i++)
              for(LL j=0;j<=2;j++)
                  for(LL k=0;k<=2;k++)
                      for(LL v=0;v<=2;v++)
                      {
                          if(!((k==1&&v==1)||(k==2&&v==2)))
                          dp[l][r][i][j]=((dp[l][loser][i][k]*dp[loser+1][r][v][j])%mod+dp[l][r][i][j]); for(LL i=0;i<=2;i++)            for(LL j=0;j<=2;j++)                for(LL k=0;k<=2;k++)                    for(LL v=0;v<=2;v++)                    {                        if(!((k==1&&v==1)||(k==2&&v==2)))                        dp[l][r][i][j]=((dp[l][loser][i][k]*dp[loser+1][r][v][j])%mod+dp[l][r][i][j]);

  然后dfs记忆化就行了.

#include
#define LL long long
using namespace std;
LL dp[744][744][3][3];
LL b[703];
LL love[703];
LL num=0;
const LL mod = 1000000007;
LL n;
void st()
{
    stacks;
    for(LL i=1;i<=n;i++)
    {
        if(b[i]==1) s.push(i);
        else 
        {
            love[s.top()]=i;
            love[i]=s.top();
            s.pop();
        }
    }
}
void dfs(LL l,LL r)
{
    if(l==r-1) 
    {
        dp[l][r][0][1]=dp[l][r][0][2]=dp[l][r][1][0]=dp[l][r][2][0]=1;
        return;
    }
    if(love[l]==r)
    {
        dfs(l+1,r-1);
        for(LL i=0;i<=2;i++)
            for(LL j=0;j<=2;j++)
            {
                if(i!=1)
                    dp[l][r][1][0]=(dp[l][r][1][0]+dp[l+1][r-1][i][j])%mod;
                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!=2)
                    dp[l][r][2][0]=(dp[l][r][2][0]+dp[l+1][r-1][i][j])%mod;
            }
        return;
    }
    else
    {
        LL loser=love[l];
        dfs(l,loser);
        dfs(loser+1,r);
        for(LL i=0;i<=2;i++)
            for(LL j=0;j<=2;j++)
                for(LL k=0;k<=2;k++)
                    for(LL v=0;v<=2;v++)
                    {
                        if(!((k==1&&v==1)||(k==2&&v==2)))
                        dp[l][r][i][j]=((dp[l][loser][i][k]*dp[loser+1][r][v][j])%mod+dp[l][r][i][j])%mod;
                    }
        return;
    }
}
int main()
{
    string a;
    cin>>a;
    for(LL i=0;iif(a[i]=='(')
            b[i+1]=1;
        else
            b[i+1]=0;
    }
    n=a.size();
    st();
    dfs(1,n);
    LL ans=0;
    for(LL i=0;i<=2;i++)
        for(LL j=0;j<=2;j++)
        {
            ans+=dp[1][n][i][j]%mod;
            ans%=mod;
        }
    cout<