CodeForces 990C

Posted 2020-12-28 daybreaking

Description

A bracket sequence is a string containing only characters "(" and ")".

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You are given $\mathrm{nn bracket\; sequences s1,s2,\dots ,sns1,s2,\dots ,sn.\; Calculate\; the\; number\; of\; pairs i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.\; Operation ++ means\; concatenation\; i.e.\; "()("\; +\; ")()"\; =\; "()()()".}$

If ${\mathrm{si+sjsi+sj and sj+sisj+si are\; regular\; bracket\; sequences\; and i\ne ji\ne j,\; then\; both\; pairs (i,j)(i,j) and (j,i)(j,i) must\; be\; counted\; in\; the\; answer.\; Also,\; if si+sisi+si is\; a\; regular\; bracket\; sequence,\; the\; pair (i,i)(i,i) must\; be\; counted\; in\; the\; answer.}}^{}$

Input

The first line contains one integer $\mathrm{n(1\le n\le 3?105)n(1\le n\le 3?105) —\; the\; number\; of\; bracket\; sequences.\; The\; following nn lines\; contain\; bracket\; sequences\; — non-empty strings\; consisting\; only\; of\; characters\; "("\; and\; ")".\; The\; sum\; of\; lengths\; of\; all\; bracket\; sequences\; does\; not\; exceed 3?1053?105.}$

Output

In the single line print a single integer — the number of pairs $\mathrm{i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.}$

Sample Input

Input

3
)
()
(

Output

2

Input

2
()
()

Output

4

Hint

In the first example, suitable pairs are $(3,1)(3,1) and (2,2)(2,2).$

In the second example, any pair is suitable, namely $(1,1),(1,2),(2,1),(2,2)(1,1),(1,2),(2,1),(2,2).$

 

括号匹配问题，需要注意几个细节。

有以下几种情况是不能与其他组括号构成合法的括号：)(         )(((      )))(

已经是合法的括号组不需要与不合法的括号组匹配。

 

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<stack>
 8 #include<deque>
 9 #include<map>
10 #include<iostream>
11 using namespace std;
12 typedef long long  LL;
13 const double pi=acos(-1.0);
14 const double e=exp(1);
15 const int N = 300010;
16 
17 #define lson i << 1,l,m
18 #define rson i << 1 | 1,m + 1,r
19 
20 map<LL,LL> mp;
21 char con[N];
22 int main()
23 {
24     LL i,p,j,n,check;
25     LL cont=0,ans=0,len1,len2;
26     scanf("%lld",&n);
27     getchar();
28     for(j=1;j<=n;j++)
29     {
30         p=check=0;
31         len1=len2=0;
32         memset(con,0,sizeof(0));
33         scanf("%s",con);
34         for(i=0;i<300009;i++)
35         {
36             if(con[i]==0)
37                 break;
38             if(con[i]==‘(‘)
39             {
40                 len1++;
41                 p++;
42             }
43             else
44             {
45                 p--;
46                 if(len1)
47                     len1--;
48                 else
49                     len2++;
50             }
51         }
52         if(len1==0&&len2==0)
53             cont++;
54         else
55         {
56             if(len1==0)
57                 mp[p]++;
58             if(len2==0)
59                 mp[p]++;
60         }
61     }
62     ans=cont*cont;
63 
64     map<LL,LL>::iterator it1;
65     for(it1=mp.begin();it1!=mp.end();it1++)
66     {
67         if(it1->first>0)
68             break;
69         if(mp[-(it1->first)]>0)
70             ans+=(it1->second)*mp[-(it1->first)];
71     }
72     printf("%lld
",ans);
73     return 0;
74 }

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