E. Arthur and Brackets
time limit per test

2 seconds

memory limit per test

128 megabytes

input

standard input

output

standard output

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 test(s)
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
(())()
#include<iostream>
#include<algorithm>
#include<vector> using namespace std; int l[], r[];
vector<int> v;
string s; int main()
{
int n;
cin >> n;
for(int i = ; i < n; i++)
{
cin >> l[i] >> r[i];
s += '(';
v.push_back(i);
while(v.size() && l[v.back()] <= )
{
int x = v.back();
if(r[x] < )
{
cout << "IMPOSSIBLE" << endl;
return ;
}
for(int i = ; i < v.size(); i++)
l[v[i]] -= , r[v[i]] -= ;
s += ')';
v.pop_back();
}
}
if(v.size())
cout << "IMPOSSIBLE" << endl;
else
cout << s << endl;
return ;
}

Codeforces Round #288 (Div. 2) E. Arthur and Brackets 贪心的更多相关文章

  1. Codeforces Round #288 (Div. 2) E. Arthur and Brackets [dp 贪心]

    E. Arthur and Brackets time limit per test 2 seconds memory limit per test 128 megabytes input stand ...

  2. Codeforces Round #288 (Div. 2) E. Arthur and Brackets

    题目链接:http://codeforces.com/contest/508/problem/E 输入一个n,表示有这么多对括号,然后有n行,每行输入一个区间,第i行的区间表示从前往后第i对括号的左括 ...

  3. 贪心+模拟 Codeforces Round #288 (Div. 2) C. Anya and Ghosts

    题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, ...

  4. 贪心 Codeforces Round #288 (Div. 2) B. Anton and currency you all know

    题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再 ...

  5. Codeforces Round #297 (Div. 2)D. Arthur and Walls 暴力搜索

    Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx ...

  6. BFS Codeforces Round #297 (Div. 2) D. Arthur and Walls

    题目传送门 /* 题意:问最少替换'*'为'.',使得'.'连通的都是矩形 BFS:搜索想法很奇妙,先把'.'的入队,然后对于每个'.'八个方向寻找 在2*2的方格里,若只有一个是'*',那么它一定要 ...

  7. Codeforces Round #297 (Div. 2)C. Ilya and Sticks 贪心

    Codeforces Round #297 (Div. 2)C. Ilya and Sticks Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  8. Codeforces Round #288 (Div. 2)

    A. Pasha and Pixels     题意就是给一个n*m的矩阵,k次操作,一开始矩阵全白,一次操作可以染黑一个格子,问第几次操作可以使得矩阵中存在一个2*2的黑色矩阵.直接模拟即可 代码: ...

  9. Codeforces Round #311 (Div. 2) C. Arthur and Table Multiset

    C. Arthur and Table Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/557/p ...

随机推荐

  1. AndroidManifest.xml权限设置

      访问登记属性 android.permission.ACCESS_CHECKIN_PROPERTIES ,读取或写入登记check-in数据库属性表的权限 获取错略位置 android.permi ...

  2. js实现table导出Excel,保留table样式

    浏览器环境:谷歌浏览器 1.在导出Excel的时候,保存table的样式,有2种方法,①是在table的行内写style样式,②是在模板里面添加样式 2.第一种方式:行内添加样式 <td sty ...

  3. git忽略特殊文件或文件夹

    1.在项目目录中添加“.gitignore”文件,项目目录就是你存放git工程的目录就是有“.git”目录的目录 vi .gitignore 2.在文件中添加如下内容,其中“/runtime/”是忽略 ...

  4. 洛谷P2024食物链

    传送门啦 这道题的特殊之处在于对于任意一个并查集,只要告诉你某个节点的物种,你就可以知道所有节点对应的物种. 比如一条长为4的链 甲->乙->丙->丁 ,我们知道乙是A物种.那么甲一 ...

  5. 洛谷P1841重要的城市

    传送门啦 重要城市有三个性质如下: 1.重要城市能对其他两个不同城市的最短路径做出贡献 2.重要城市具有唯一性,如果两不同城市之间的最短路径有两种中间城市情况,那么这两个中间城市可以彼此代替,就都不能 ...

  6. 可怕的线程上下文类装载器(TCCL)

    在明天的 OSGi 2012 社区活动上,我将以"如何使你的类库在不依赖 OSGi 的情况下进行友好地 OSGi"为主题进行演讲.在演讲中我将会提及 Java 的线程上下文类加载器 ...

  7. KnockoutJs学习笔记(四)

    由于Writable computed observables和How dependency tracking works的要求相对较高,我先跳过这两篇,学习Pure computed observa ...

  8. Observer 观察者

    意图 定义对象间的一种一对多的依赖关系 ,当一个对象的状态发生改变时, 所有依赖于它的对象都得到通知并被自动更新. 动机 一致性,松耦合 需要维护相关对象间的一致性.我们不希望为了维持一致性而使各类紧 ...

  9. 【LOJ】#2275. 「JXOI2017」颜色

    题解 我们枚举右端点判断合法的左端点有哪些 首先,记录一下右端点右边的点的pre,也就是这个数字前一个出现的位置,取所有小于枚举右端点r的值中最大的一个做为l,用优先队列维护即可,[l + 1,r]就 ...

  10. Just a Hook (线段树)

    给你n个数(初始时每个数的值为1),m个操作,每个操作把区间[l,r]里的数更新为c,问最后这n个数的和是多少. 区域更新用懒惰标记 #include<bits/stdc++.h> usi ...