题意：

在这里插入图片描述

思路：

考虑对于线段，如何建模。
我们考虑先将线段转换成左闭右开的形式，将左右点连起来。
再考虑每个点，将所有离散化后的点拿出来，每个点都有一个度，现在问题就是给每个边定向，让入度和出度之差 $≤1\le1$ 即可。
这就是欧拉回路的经典问题了，将奇度点之间连边即可。

// Problem: E. Points and Segments
// Contest: Codeforces - Codeforces Round #245 (Div. 1)
// URL: https://codeforces.com/contest/429/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=2000010,M=N,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m;
vector<int>v;
PII p[N];
int type;
int h[N], e[M], ne[M], idx;
bool used[M];
int ans[M], cnt;
int d[N];void add(int a, int b)
{e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}void dfs(int u)
{for (int &i = h[u]; ~i;){if (used[i]){i = ne[i];continue;}used[i] = true;if (type == 1) used[i ^ 1] = true;int t;if (type == 1){t = i / 2 + 1;if (i & 1) t = -t;}else t = i + 1;int j = e[i];i = ne[i];if(t>0) ans[t]=1;else ans[-t]=0;dfs(j);// if(t>0) ans[t]=1;// else ans[-t]=0;// cout<<t<<endl;}
}int find(int x) {return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);memset(h,-1,sizeof(h));type=1;scanf("%d",&n);for(int i=1;i<=n;i++) {int l,r; scanf("%d%d",&l,&r);p[i]={l,r+1}; v.pb(l); v.pb(r+1);}sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());for(int i=1;i<=n;i++) {p[i].X=find(p[i].X),p[i].Y=find(p[i].Y);add(p[i].X,p[i].Y); add(p[i].Y,p[i].X);d[p[i].X]++; d[p[i].Y]++;// cout<<p[i].X<<' '<<p[i].Y<<endl;}vector<int>now;for(int i=1;i<=v.size();i++) if(d[i]%2==1) now.pb(i);for(int i=0;i<now.size();i+=2) add(now[i],now[i+1]),add(now[i+1],now[i]);for(int i=1;i<=v.size();i++) if(h[i]!=-1) {dfs(i);}for(int i=1;i<=n;i++) printf("%d ",ans[i]);return 0;
}
/**/