题意：

在这里插入图片描述

思路：

欧拉回路经典题。
将其转换成图上问题，对于横纵坐标我们将其分开，对于 $(x, y)$ 我们将其横纵坐标连一个无向边，现在问题就转换成了我们需要对每条边定向，使得点的入度和出度之差 $≤1\le1$ 。当然如果有奇度顶点，我们将奇度点之间连接即可，最终输出的时候去掉这些边。
注意图可能不连通，需要判断每个点来跑。

// Problem: A - Mike and Fish
// Contest: Virtual Judge - Camp 2
// URL: https://vjudge.net/contest/454780#problem/A
// Memory Limit: 262 MB
// Time Limit: 3000 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,H=2e5;
const double eps=1e-6;int type;
int n, m;
int h[N], e[M], ne[M], idx;
bool used[M];
int ans[M], cnt;
int d[N],a[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];dfs(j);ans[ ++ cnt] = t;}
}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 a,b; scanf("%d%d",&a,&b);d[a]++; d[b+H]++;add(a,b+H); add(b+H,a);}vector<int>v;for(int i=1;i<=4e5;i++) if(d[i]%2==1) v.pb(i);assert(v.size()%2==0);for(int i=0;i<v.size();i+=2) add(v[i],v[i+1]),add(v[i+1],v[i]);for(int i=1;i<=4e5;i++) if(h[i]!=-1) {dfs(i);}for(int i=1;i<=cnt;i++) {if(abs(ans[i])>n) continue;if(ans[i]<0) a[-ans[i]]=1;else a[ans[i]]=0;}for(int i=1;i<=n;i++) printf("%c",a[i]==1? 'r':'b');puts("");return 0;
}
/**/