Arthur and Brackets CodeForces - 508E (贪心,括号匹配)

大意: n个括号, 位置未知, 给出每对括号左右间距, 求输出一个合法括号序列.

最后一个括号间距一定为1, 从右往左遍历, 每次括号划分越小越好.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e6+10;
int n, l[N], r[N], len[N], vis[N];
char s[N];
int fa[N];
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d%d", l+i,r+i);
	REP(i,1,n+4) fa[i]=i;
	if (l[n]>1) return puts("IMPOSSIBLE"),0;
	len[n] = 2;
	PER(i,1,n-1) {
		int t = 1;
		if (l[i]==1) {len[i]=2;continue;}
		for (int j=find(i+1); j<=n; j=find(j)) {
			t += len[j];
			fa[j] = j+1;
			if (l[i]<=t&&t<=r[i]) {
				len[i] = t+1;
				break;
			}
		}
		if (!len[i]) return puts("IMPOSSIBLE"),0;
	}
	int now = 1;
	REP(i,1,n) {
		while (s[now]) ++now;
		s[now] = '(', s[now+len[i]-1] = ')';
	}
	puts(s+1);
}