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  • Codeforces Round #501 (Div. 3) F. Bracket Substring

    https://codeforces.com/problemset/problem/1015/F

    dp

    '求包含某个子串的个数' 类型

    kmp

        ///it is advised that all character begins at index 1

     1 #include <cstdio>
     2 #include <cstdlib>
     3 #include <cmath>
     4 #include <cstring>
     5 #include <string>
     6 #include <algorithm>
     7 #include <iostream>
     8 using namespace std;
     9 #define ll long long
    10 
    11 const double eps=1e-8;
    12 const ll inf=1e9;
    13 const ll mod=1e9+7;
    14 const int maxn=2e2+10;
    15 
    16 char sstr[maxn];
    17 
    18 int str[maxn],pre[maxn],nex[maxn][2];
    19 ll f[maxn][maxn][maxn][2];
    20 
    21 int main()
    22 {
    23     int n,nn,len,i,j,k,l,m,jj,lll;
    24     scanf("%d",&n);
    25     nn=n<<1;
    26     ///it is advised that all character begins at index 1
    27     scanf("%s",sstr+1);
    28     len=strlen(sstr+1);
    29     for (i=1;i<=len;i++)
    30         if (sstr[i]=='(')
    31             str[i]=0;
    32         else
    33             str[i]=1;
    34 
    35     ///kmp makes O(1) visit
    36     j=0;
    37     pre[1]=0;
    38     for (i=2;i<=len;i++)
    39     {
    40         while (j!=0 && str[j+1]!=str[i])
    41             j=pre[j];
    42         if (str[j+1]==str[i])
    43             j++;
    44         pre[i]=j;
    45     }
    46 
    47     for (i=0;i<=len-1;i++)
    48         for (j=0;j<=1;j++)
    49             if (str[i+1]==j)
    50                 nex[i][j]=i+1;
    51             else
    52                 nex[i][j]=nex[pre[i]][j];
    53 
    54     f[0][0][0][0]=1;
    55     ///pos ith
    56     for (i=1;i<=nn;i++)
    57         ///j '('
    58         for (j=0;j<=min(i-1,n);j++)
    59             ///current char
    60             for (k=0;k<=1;k++)
    61             {
    62                 jj=j+1+(-2)*(k==1);
    63                 if (jj==-1)
    64                     continue;
    65                 (f[i][jj][0][1]+=f[i-1][j][0][1])%=mod;
    66                 ///the lth str
    67                 for (l=0;l<=min(len-1,i-1);l++)
    68                 {
    69                     lll=nex[l][k];
    70                     if (lll==len)
    71                         (f[i][jj][0][1]+=f[i-1][j][l][0])%=mod;
    72                     else
    73                         (f[i][jj][lll][0]+=f[i-1][j][l][0])%=mod;
    74                 }
    75             }
    76     printf("%lld",f[nn][0][0][1]);
    77     return 0;
    78 }
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  • 原文地址:https://www.cnblogs.com/cmyg/p/11110273.html
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