「CF508E」Arthur and Brackets

2015年1月28日3,7870

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 [l_i, r_i], 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 [l_i, r_i].

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 l_i and r_i (1 ≤ l_i ≤ r_i < 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

1 1

output

()()()()

()()()()

input

5 5

3 3

1 1

output

((()))

((()))

input

5 5

3 3

2 2

output

IMPOSSIBLE

1	IMPOSSIBLE

input

2 3

1 4

output

(())()

(())()

题解

经过尝试构造后发现。。。

将括号的dis设小不会更劣

如（（））变成（）（）更有利于加入其它的括号

于是从后往前依次考虑每个括号，使得它尽量包含少的括号

复杂度n^2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<set>
#include<ctime>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#define eps 1e-8
#define inf 1000000000
#define pa pair<int,int>
#define ll long long 
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
vector<int> st;
int ans[605],q[1205];
int l[605],r[605];
bool mark[605];
void print()
{
	int now=1;
	for(int i=1;i<=n;i++)
	{
		while(q[now])now++;
		q[now]=1;q[now+ans[i]]=-1;
	}
	for(int i=1;i<=2*n;i++)
		if(q[i]==1)printf("(");
		else printf(")");
}
bool solve(int t)
{
	st.clear();
	int tot=0;
	if(l[t]==1){ans[t]=1;return 1;}
	if(ans[t])return 1;
	for(int j=t+1;j<=n&&tot+1<l[t];j++)
		if(ans[j])
		{
			if(!mark[j]){tot+=ans[j]+1;st.push_back(j);}
		}
		else return 1;
	tot++;
	if(tot<l[t]||tot>r[t])return 0;
	for(int i=0;i<st.size();i++)
		mark[st[i]]=1;
	ans[t]=tot;
	return 1;
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		l[i]=read(),r[i]=read();
		if(l[i]%2==0)l[i]++;
	}
	for(int x=n;x;x--)
		if(!solve(x)){puts("IMPOSSIBLE");return 0;}
	print();
	return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<set>

#include<ctime>

#include<vector>

#include<queue>

#include<algorithm>

#include<map>

#include<cmath>

#define eps 1e-8

#define inf 1000000000

#define pa pair<int,int>

#define ll long long

using namespace std;

int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n;

vector<int> st;

int ans[605],q[1205];

int l[605],r[605];

bool mark[605];

void print()

{

int now=1;

for(int i=1;i<=n;i++)

{

while(q[now])now++;

q[now]=1;q[now+ans[i]]=-1;

}

for(int i=1;i<=2*n;i++)

if(q[i]==1)printf("(");

else printf(")");

}

bool solve(int t)

{

st.clear();

int tot=0;

if(l[t]==1){ans[t]=1;return 1;}

if(ans[t])return 1;

for(int j=t+1;j<=n&&tot+1<l[t];j++)

if(ans[j])

{

if(!mark[j]){tot+=ans[j]+1;st.push_back(j);}

}

else return 1;

tot++;

if(tot<l[t]||tot>r[t])return 0;

for(int i=0;i<st.size();i++)

mark[st[i]]=1;

ans[t]=tot;

return 1;

}

int main()

{

n=read();

for(int i=1;i<=n;i++)

{

l[i]=read(),r[i]=read();

if(l[i]%2==0)l[i]++;

}

for(int x=n;x;x--)

if(!solve(x)){puts("IMPOSSIBLE");return 0;}

print();

return 0;

}

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