CF思维联系– Codeforces-990C Bracket Sequences Concatenation Problem（括号匹配+模拟）

风骨散人Chiam

发布于 2020-10-29 09:44:45

3690

发布于 2020-10-29 09:44:45

文章被收录于专栏：CSDN旧文

ACM思维题训练集合 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 n bracket sequences s1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n) such that the bracket sequence si+sj is a regular bracket sequence. Operation + means concatenation i.e. “()(” + “)()” = “()()()”.

If si+sj and sj+si are regular bracket sequences and i≠j, then both pairs (i,j) and (j,i) must be counted in the answer. Also, if si+si is a regular bracket sequence, the pair (i,i) must be counted in the answer.

Input The first line contains one integer n(1≤n≤3⋅105) — the number of bracket sequences. The following n lines contain bracket sequences — non-empty strings consisting only of characters “(” and “)”. The sum of lengths of all bracket sequences does not exceed 3⋅105.

Output In the single line print a single integer — the number of pairs i,j(1≤i,j≤n) such that the bracket sequence si+sj is a regular bracket sequence.

Examples Input 3 ) () ( Output 2 Input 2 () () Output 4 Note In the first example, suitable pairs are (3,1) and (2,2).

In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2). 模拟稍微有一下就可以了

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    t f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}

#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 1005
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
map<LL, LL> mp;
char con[N];
int main()
{
    LL i, p, j, n, check;
    LL cont = 0, ans = 0, len1, len2;
    scanf("%lld", &n);
    getchar();
    for (j = 1; j <= n; j++)
    {
        p = check = 0;
        len1 = len2 = 0;
        memset(con, 0, sizeof(0));
        scanf("%s", con);
        for (i = 0; i < 300009; i++)
        {
            if (con[i] == 0)
                break;
            if (con[i] == '(')
            {
                len1++;
                p++;
            }
            else
            {
                p--;
                if (len1)
                    len1--;
                else
                    len2++;
            }
        }
        if (len1 == 0 && len2 == 0)
            cont++;
        else
        {
            if (len1 == 0)
                mp[p]++;
            if (len2 == 0)
                mp[p]++;
        }
    }
    ans = cont * cont;

    map<LL, LL>::iterator it1;
    for (it1 = mp.begin(); it1 != mp.end(); it1++)
    {
        if (it1->first > 0)
            break;
        if (mp[-(it1->first)] > 0)
            ans += (it1->second) * mp[-(it1->first)];
    }
    printf("%lld\n", ans);
    return 0;
}

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原始发表：2020/03/01 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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