CF - div2 - 459 - C - The Monster_camedivc-2

C. The Monster

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn them on and off with his mind). He can’t directly tell his mom where he is, because the monster that took him to the Upside Down will know and relocate him.

Thus, he came up with a puzzle to tell his mom his coordinates. His coordinates are the answer to the following problem.

A string consisting only of parentheses (‘(’ and ‘)’) is called a bracket sequence. Some bracket sequence are called correct bracket sequences. More formally:

Empty string is a correct bracket sequence.
if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.
A string consisting of parentheses and question marks (‘?’) is called pretty if and only if there’s a way to replace each question mark with either ‘(’ or ‘)’ such that the resulting string is a non-empty correct bracket sequence.

Will gave his mom a string s consisting of parentheses and question marks (using Morse code through the lights) and his coordinates are the number of pairs of integers (l, r) such that 1 ≤ l ≤ r ≤ |s| and the string slsl + 1… sr is pretty, where si is i-th character of s.

Joyce doesn’t know anything about bracket sequences, so she asked for your help.

Input
The first and only line of input contains string s, consisting only of characters ‘(‘, ‘)’ and ‘?’ (2 ≤ |s| ≤ 5000).

Output
Print the answer to Will’s puzzle in the first and only line of output.

Examples
input
((?))
output
4
input
??()??
output
7
Note
For the first sample testcase, the pretty substrings of s are:

1.”(?” which can be transformed to “()”.
2.”?)” which can be transformed to “()”.
3.”((?)” which can be transformed to “(())”.
4.”(?))” which can be transformed to “(())”.
For the second sample testcase, the pretty substrings of s are:

1.”??” which can be transformed to “()”.
2.”()”.
3.”??()” which can be transformed to “()()”.
4.”?()?” which can be transformed to “(())”.
5.”??” which can be transformed to “()”.
6.”()??” which can be transformed to “()()”.
7.”??()??” which can be transformed to “()()()”.

题目链接：http://codeforces.com/contest/918/problem/C


题意：

• 输入一个字符串 s

问：有多少个子串是好串？

好串的定义：

1. Empty string is a correct bracket sequence.
2. if s is a correct bracket sequence, then (s) is also a correct bracket sequence.
3. if s and t are correct bracket sequences, then st (concatenation of s and t) is also a correct bracket sequence.

解题思路:见代码

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
using namespace std;
typedef long long LL;

const int maxn = 50005;
int l[maxn];//l[i]:记录的是以坐标 i 为左端点，能满足为好串，并使得串最长的【右端点坐标】 + 1 的坐标。
int r[maxn];//r[i]:记录的是以坐标 i 为右端点，能满足为好串，并使得串最长的【左端点坐标】 - 1 的坐标。
char s[maxn];

int main(){
    cin >> s + 1;
    int len = strlen(s + 1);
    for(int i = 1;i <= len;i++){//从前往后
        int cnt = 0,pos = len + 1;
        for(int j = i;j <= len;j++){
            if(s[j] == '(' || s[j] == '?') cnt++;
            else cnt--;
            if(cnt < 0){
                pos = j;
                break;
            }
        }
        l[i] = pos;
    }
    for(int i = len;i >= 1;i--){//从后往前
        int cnt = 0,pos = 0;
        for(int j = i;j >= 1;j--){
            if(s[j] == ')' || s[j] == '?') cnt++;
            else cnt--;
            if(cnt < 0){
                pos = j;
                break;
            }
        }
        r[i] = pos;
    }
    int sum = 0;
    for(int i = 1;i <= len;i++){//枚举区间(i,j)
        int a = 0,b = 0,c = 0;
        for(int j = i;j <= len;j++){
            if(l[i] <= j) break;//以坐标 i 为左端点，能满足为好串，并使得串最长的【右端点坐标】 + 1 的坐标 小于等于 j。说明以坐标 i 为左端点的好串不存在。
            if(s[j] == '(') a++;
            else if(s[j] == ')') b++;
            else if(s[j] == '?') c++;
            if((abs(a - b) + c) % 2 == 0){//abs(a - b)代表还缺 "(" 或 ")" 的个数，若缺少的能用 "?" 补全 
                if(r[j] < i) sum++;//且以坐标 i 为右端点，能满足为好串，并使得串最长的【左端点坐标】 - 1 的坐标 小于 i 的话，说明区间(i,j)是一个好串。
            }
        }
    }
    cout << sum << endl;
    return 0;
}
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