涉及知识点：图的遍历。

我们观察样例可以发现，染色之后的图是一颗树，而且还是 dfs 树。

题目要求所以路径上的颜色都是交替的，所以直接交替染色即可。

注意：建图的时候需要记录当前边的编号。

代码

#include <bits/stdc++.h>
#define int long long
#define ll __int128
#define db double
#define ldb long double
#define vo void
#define endl '\n'
#define il inline
#define re register
#define ve vector
#define p_q priority_queue
#define PII pair<int, int>
using namespace std;

//#define O2 1
#ifdef O2
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#endif

namespace OI {
	template <typename T>
	il T read() {
		T x = 0, f = 1;
		int ch = getchar();
		while (!isdigit(ch)) {
			if (ch == '-') f = -1;
			ch = getchar();
		}
		while (isdigit(ch)) {
			x = (x << 3) + (x << 1) + (ch ^ 48);
			ch = getchar();
		}
		return x * f;
	}
	template <typename TE>
	il void write(TE x) {
		if (x < 0) {
			x = -x;
			putchar('-');
		}
		TE sta[35];
		int top = 0;
		do {
			sta[top++] = x % 10, x /= 10;
		} while (x);
		while (top) putchar(sta[--top] + '0');
	}
	il string read_with_string() {
		string s = "";
		char ch = getchar();
		while ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z') || (ch >= '0' && ch <= '9')) {
			s += ch;
			ch = getchar();
		}
		return s;
	}
	il void write_with_string(string s) {
		for (int i = 0; i < s.size(); i ++ ) putchar(s[i]);
	}
}
namespace COMB {
	int fact[200000];
	int Triangle[1010][1010];
	void Fact(int n, int mod) {
		fact[0] = 1;
		for (int i = 1; i <= n; i ++ ) fact[i] = ((fact[i - 1]) % mod * (i % mod)) % mod;
	}
	void Pascal_s_triangle(int n, int mod) {
		for (int i = 0; i <= n; i ++ ) Triangle[i][0] = 1;
		for (int i = 1; i <= n; i ++ )
			for (int j = 1; j <= i; j ++ )
				Triangle[i][j] = (Triangle[i - 1][j] + Triangle[i - 1][j - 1]) % mod;
	}
	int pw(int x, int y, int mod) {
		int res = 1;
		while (y) {
			if (y & 1) res = ((res % mod) * (x % mod)) % mod;
			x = (x % mod) * (x % mod) % mod;
			y >>= 1;
		}
		return res;
	}
	int pw(int x, int y) {
		int res = 1;
		while (y) {
			if (y & 1) res *= x;
			x *= x;
			y >>= 1;
		}
	}
	int GCD(int x, int y, int mod) {
		return __gcd(x, y) % mod;
	}
	int LCM(int x, int y, int mod) {
		return (((x % mod) * (y % mod)) % mod / (GCD(x, y, mod) % mod)) % mod;
	}
	int C(int n, int m, int mod) {
		if (m > n || m < 0) return 0;
		return fact[n] * pw(fact[m], mod - 2, mod) % mod * pw(fact[n - m], mod - 2, mod) % mod;
	}
	int Ask_triangle(int x, int y) {
		return Triangle[x][y];
	}
}
using namespace OI;
using namespace COMB;

//#define fre 1
#define IOS 1
//#define multitest 1

const int N = 2e5 + 10;
const int M = 4e5 + 10;
const int inf = 1e12;

int n, m;
int color[N];
int f[N];
ve<PII> g[N];

il void Init() {
	memset(color, -1, sizeof color);
	cin >> n >> m;
	for (int i = 1; i <= m; i ++ ) {
		int x, y;
		cin >> x >> y;
		g[x].push_back({y, i});
		g[y].push_back({x, i});
	}
}

il void dfs(int u, int cl) {
	f[u] = 1;
	for (int i = 0; i < g[u].size(); i ++ ) {
		int v = g[u][i].first;
		if (f[v]) continue;
		color[g[u][i].second] = cl % 2;
		dfs(v, cl + 1);
	}
}

il void Solve() {
	for (int i = 1; i <= n; i ++ )
		if (!f[i])
			dfs(i, 0);
	for (int i = 1; i <= m; i ++ ) {
		if (color[i] == -1) cout << "G";
		else if (color[i] == 0) cout << "B";
		else cout << "R";
	}
}

signed main() {
	int T;
#ifdef IOS
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
#ifdef fre
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
#ifdef multitest
	cin >> T;
#else
	T = 1;
#endif
	while (T--) {
		Init();
		Solve();
	}
	return 0;
}
/*

*/