CodeForces 508E Arthur and Brackets 贪心

 题目：

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).

Examples

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

(())()

题意：

按从左到右的顺序给出n对括号的距离范围[l,r]，问这些括号是否能组成合法的括号对。

样例二：

题解：

最右边的一对括号距离肯定要是1，因为它右边不会再有括号被包含了，

同理，从右往左扫，扫到第i对括号的时候，它的右边的所有括号的长度都是确定下来的，然后从i+1到n，考虑最小的能够被i合法包含的括号对，这里贪心找最小能够为左边提供更多方案，从只要有解，都不会漏掉。

时间复杂度总共O(n^2)。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;

const int maxn = 666;

int n;

struct Node {
    int x, y;
    bool operator < (const Node& tmp) const {
        return x>tmp.x || (x == tmp.x) && y>tmp.y;
    }
}nds[maxn];

char ans[maxn * 2];
int used[maxn * 2];
int fst, las;

void init() {
    memset(used, 0, sizeof(used));
}

int main() {
    //    freopen("data_in.txt","r",stdin);
    while (scanf("%d", &n) == 1 && n) {
        init();
        for (int i = 0; i<n; i++) {
            scanf("%d%d", &nds[i].x, &nds[i].y);
        }
        int su = 1;
        for (int i = n - 1; i >= 0; i--) {
            if (nds[i].x == 1) {
                nds[i].x = 1;
            }
            else {
                int sum = 1, flag = 0;
                for (int j = i + 1; j<n;) {
                    sum += (nds[j].x + 1);
                    if (sum >= nds[i].x&&sum <= nds[i].y) {
                        nds[i].x = sum; flag = 1; break;
                    }
                    j += (nds[j].x + 1) / 2;
                }
                if (flag == 0) {
                    su = 0; break;
                }
            }
        }
        if (!su) printf("IMPOSSIBLE\n");
        else {
            for (int i = 0; i<n; i++) {
                for (int j = 0; j<2 * n; j++) {
                    if (used[j] == 0) {
                        ans[j] = '('; used[j] = 1;
                        ans[j + nds[i].x] = ')'; used[j + nds[i].x] = 1;
                        break;
                    }
                }
            }
            for (int i = 0; i<2 * n; i++) printf("%c", ans[i]);
            printf("\n");
        }
    }
    return 0;
}