SGU179 Brackets light

179. Brackets light

time limit per test: 0.25 sec. 
memory limit per test: 131072 KB

input: standard 
output: standard

There is a correct brackets sequence. It's length doesn't exceed 10000 symbols. 
Your task is to find next (in lexicographic order) correct brackets sequence with the same length. You may assume that '(' < ')'.

Input

The first line of the input contains correct brackets sequence. There are only '(' and ')' symbols in the input.

Output

Write sought sequence in the single line of the output or 'No solution' if solution doesn't exist.

Sample test(s)

Input

(())()

Output

()(())

题目大意：给你一个括号序列，规定(小于),求当前括号序列的下一个字典序的合法的括号序列.

分析：比较麻烦的是要求合法.生成下一个字典序的括号序列可以用类似生成下一个全排列的思想：每次从后往前找第一个单调上升的位置，那么后面的就都是单调下降的了.把这个位置的数和它前面的那一个数互换，再把它后面的所有数翻转，这就是具体的思想.那么如何满足合法的要求呢？不断地求下一个序列，直到满足要求为止.那么这样就有一个问题了，怎么判断一个序列一定都够变成有解的呢？找特征！把不能满足要求的序列打个表，就可以发现如果初始序列长成这样：()()()(),那么是无解的，剩下的情况全都有解.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

char s[10010];
int len, cnt;
bool flag = false;

void update()
{
    for (int i = len; i >= 1; i--)
        if (s[i] > s[i - 1])
        {
            swap(s[i], s[i - 1]);
            reverse(s + i + 1, s + len + 1);
            break;
        }
}

bool check()
{
    cnt = 0;
    for (int i = 1; i <= len; i++)
    {
        if (s[i] == '(')
            cnt++;
        else
        {
            if (cnt == 0)
                return false;
            else
                cnt--;
        }
    }
    if (cnt)
        return false;
    return true;
}

int main()
{
    while (scanf("%s", s + 1) != EOF)
    {
        len = strlen(s + 1);
        flag = false;
        for (int i = 1; i <= len; i++)
            if (s[i] == '(' && s[i + 1] != ')')
            {
                flag = true;
                break;
            }
        if (!flag)
            puts("No solution");
        else
        {
            while (1)
            {
                update();
                if (check())
                    break;
            }
            for (int i = 1; i <= len; i++)
                printf("%c", s[i]);
            printf("\n");
        }
    }

    return 0;
}