code forces 990C
2 seconds
256 megabytes
standard input
standard output
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 nn bracket sequences s1,s2,…,sns1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence. Operation ++ means concatenation i.e. "()(" + ")()" = "()()()".
If si+sjsi+sj and sj+sisj+si are regular bracket sequences and i≠ji≠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.
The first line contains one integer n(1≤n≤3⋅105)n(1≤n≤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.
In the single line print a single integer — the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence.
3
)
()
(
2
2
()
()
4
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).
题意,寻找合法的括号匹配对数(1,3)和(3,1)不同
题解:感谢llw大佬讲题,思路是将每一组括号转化成对应的值来匹配,计算可匹配的值,相加即可得到答案
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+;
typedef long long ll;
ll n,mx=;
char str[maxn];
map<ll,ll> mp;
int main(){
ios::sync_with_stdio(false);
cin>>n;
while(n--){
cin>>str;
ll flag=,k=,len=strlen(str);
for(int i=;i<len;i++){
if(str[i]=='(') k++;
else k--;
//k=0的情况是这组数据符合要求
//k>0的情况是左大于右
//k<0的情况是右大于左
if(k<) flag=min(flag,k);
}
if(flag<&&k>flag) continue; //不合法的序列不用管,譬如())(
else{
//cout<<k<<" "<<flag<<endl;
mp[k]++;
mx=max(k,mx);
}
}
ll ans=;
for(int i=;i<=mx;i++){
ans+=mp[i]*mp[-i];
}
cout<<ans<<endl;
return ;
}
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