http://codeforces.com/contest/990/problem/C
C. Bracket Sequences Concatenation Problem
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
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.
Input
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.
Output
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.
Examples
input
Copy
3)()(
output
Copy
2
input
Copy
2()()
output
Copy
4
Note
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+5;
typedef long long ll;
ll n,mx=0;
char str[maxn];
map<ll,ll> mp;
int main(){
ios::sync_with_stdio(false);
cin>>n;
while(n--){
cin>>str;
ll flag=0,k=0,len=strlen(str);
for(int i=0;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<0) flag=min(flag,k);
}
if(flag<0&&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=0;
for(int i=0;i<=mx;i++){
ans+=mp[i]*mp[-i];
}
cout<<ans<<endl;
return 0;
}
C. Bracket Sequences Concatenation Problem
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
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.
Input
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.
Output
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.
Examples
input
Copy
3)()(
output
Copy
2
input
Copy
2()()
output
Copy
4
Note
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).
原文地址：https://www.cnblogs.com/buerdepepeqi/p/9166339.html