Codeforces 508E

C - Arthur and Brackets

Description

Notice that the memory limit is non-standard.

Recently Arthur and Sasha have studied correct bracket sequences. Arthur understood this topic perfectly and become so amazed about correct bracket sequences, so he even got himself a favorite correct bracket sequence of length 2n. Unlike Arthur, Sasha understood the topic very badly, and broke Arthur’s favorite correct bracket sequence just to spite him.

All Arthur remembers about his favorite sequence is for each opening parenthesis (‘(’) the approximate distance to the corresponding closing one (‘)’). For the i-th opening bracket he remembers the segment [li, ri], containing the distance to the corresponding closing bracket.

Formally speaking, for the i-th opening bracket (in order from left to right) we know that the difference of its position and the position of the corresponding closing bracket belongs to the segment [li, ri].

Help Arthur restore his favorite correct bracket sequence!

Input

The first line contains integer n (1 ≤ n ≤ 600), the number of opening brackets in Arthur’s favorite correct bracket sequence.

Next n lines contain numbers li and ri (1 ≤ li ≤ ri < 2n), representing the segment where lies the distance from the i-th opening bracket and the corresponding closing one.

The descriptions of the segments are given in the order in which the opening brackets occur in Arthur’s favorite sequence if we list them from left to right.

Output

If it is possible to restore the correct bracket sequence by the given data, print any possible choice.

If Arthur got something wrong, and there are no sequences corresponding to the given information, print a single line “IMPOSSIBLE” (without the quotes).

Sample Input

Input

4
1 1
1 1
1 1
1 1

Output

()()()()

Input

3
5 5
3 3
1 1

Output

((()))

Input

3
5 5
3 3
2 2

Output

IMPOSSIBLE

Input

3
2 3
1 4
1 4

Output

(())()

这种带括号匹配的问题一般都能用到栈,然而我却傻乎乎的用dfs然后wa了很久。

代码借鉴了别人的，我加点注释吧orz。

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
typedef  long long LL;

#define clr(x) memset((x),0,sizeof(x))
#define inf 0x3f3f3f3f

using namespace std;
char ans[1250];
int l[650],r[650];
int pos[650];//存放第i个'('的位置
int s[1250];//栈
int top=0;
int n;
bool flag ;


int main()
{
    flag =0;
    int len=0; //表示位置
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&l[i],&r[i]);
        if(flag )  continue;
        s[++top] = i;
        pos[i] = len;
        ans[len++] = '(';
        while(top>0&& pos[s[top]]+l[s[top]] <=len)//表示要匹配栈顶左括号时候若这时候实际长度可能
        {                                          //在匹配范围之内
            if(pos[s[top]]+r[s[top]]<len)      //不在匹配范围之内
            {
                flag = 1;
                break;
            }
            else {
                ans[len++] = ')';  //匹配到，出栈
                s[top--] = 0;
            }

        }

    }
    if(!flag&&top ==0) //top!=0 还有括号没有得到匹配
    {
        puts(ans);
    }
    else
     puts("IMPOSSIBLE");
    return 0;
}