codeforce 5C Longest Regular Bracket Sequence

C. Longest Regular Bracket Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

Input

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

题目大意是给定字符串，输出 “最长的括号匹配串” 的长度和数量。对于这种括号的题目，基本都得维护一个计数器，然后根据计数器的状态去想办法。

对于给定的串，维护一个计数器cnt，遇到'('加一， 遇到')'减一；再记录一下当前照的最长括号匹配串的开始位置start。

当cnt <= 0时是需要特别注意的。

cnt 小于 0 ：说明 当前的字符无法作为 括号匹配串的起点，所以start直接跳到下一个字符，并且把cnt置为0

cnt 等于 0 ：说明从start到当前字符恰好组成了一个括号匹配串，然后检查这个括号匹配串的长度，是否需要更新结果。

上述算法中的start实际是跳跃的，每次跳跃的都是一个括号匹配串的长度。

那么对于直到结束，cnt还为正的串，最后的结果可能会有问题，例如 “(()” ，

为了解决这个问题，只要把串反过来再来一遍就行了。因为如果一个串是括号匹配的，那么该串从右往左看也必然是括号匹配的。

也就是说把“(()” 反成 “())” 

总得来说是贪心的算法,复杂度是O(n)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <stack>#include <map>#include <queue>#include <algorithm>using namespace std;void solve(char* str, int len, int& sublen, int& subnum) {    int cnt = 0;    int start = 0;    for (int i = 0; i < len; i++) {        cnt += (str[i] == '(' ? 1 : 0);        cnt -= (str[i] == ')' ? 1 : 0);        if (cnt == 0) {            if (i - start + 1 > sublen) {                sublen = i - start + 1;                subnum = 1;            } else if (i - start + 1 == sublen) {                subnum += 1;            }        } else if (cnt < 0) {            start = i + 1;            cnt = 0;        }    }}int main() {    char str1[1000005];    char str2[1000005];    scanf("%s", str1);    int len = strlen(str1);    for (int i = 0; i < len; i++) str2[i] = str1[len-i-1] == '(' ? ')' : '(';    int sublen1 = 0, subnum1 = 0, sublen2 = 0, subnum2 = 0;    solve(str1, len, sublen1, subnum1);    solve(str2, len, sublen2, subnum2);        if (sublen1 == 0 && sublen2 == 0) {        printf("0 1\n");    } else if (sublen1 < sublen2) {        printf("%d %d\n", sublen2, subnum2);    } else {        printf("%d %d\n", sublen1, subnum1);    }    return 0;}