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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
Review status
AI-generated and still unreviewed. Double-check the details before internalizing them.
Hints
Open only as much as you need to keep the solve alive.
Forget vertex IDs during the second run. Blackslex only sees a multiset/string of neighbor colors, so your whole strategy must guarantee: for every , among the colors of there is some color rule that always points to a neighbor with smaller distance to .
In a bipartite graph, every edge changes BFS distance from vertex by exactly . So each neighbor of is either on layer or layer ; there are no same-layer edges. That is doing a lot of heavy lifting here.
If the color of a vertex only depends on its BFS distance modulo , then a vertex at distance has neighbors only in layers and . Those two layer colors are distinct modulo , so the colors separate “closer” from “farther.”
Blackslex does not know , but he can infer which two adjacent-layer colors appear from the neighbor color set. If only one color appears, then all neighbors are on the same side; since and the graph is connected, that single color must include a closer neighbor, so picking any is fine.
The only tricky case is when two colors appear. With colors assigned by distance mod , the closer color is determined only by the unordered pair: for pair choose , for choose , and for choose . In letters, choose g from rg, b from gb, and r from rb.
Core Idea
Run BFS from vertex and color each vertex by .
Why this works: the graph is bipartite. For any edge , BFS distances differ by at most . Since adjacent vertices in a bipartite graph have opposite parity, the distance difference cannot be . So every edge always connects layers whose distances differ by exactly .
So if Blackslex is standing at distance , every neighbor is in layer or . One is closer, one is farther. No sideways edges, no hidden nonsense.
Coloring
Use:
rgbA vertex at distance sees neighbor colors from these possible residues:
Those two residues are always different. That means the closer neighbors and farther neighbors have different colors. Great, the graph is basically leaking the answer through modulo arithmetic.
Decoding
Blackslex only gets the colors of the neighbors, in arbitrary order.
If only one color appears, pick any neighbor of that color. Since , there is at least one shortest-path predecessor, so a closer neighbor exists. If all neighbors have the same color, the simple “pick that color” rule is fine.
If two colors appear, the missing color determines where Blackslex is modulo .
The useful table is:
r and g: choose gg and b: choose br and b: choose rIn residue form:
That is the closer layer color .
Hacked Format
Originally this is run twice:
For hacks, the input gives the graph and then actual queries with vertex plus a permutation of its sorted neighbors. So the solution can compute the coloring, simulate the color string Blackslex would see, then apply the same decoding rule.
The implementation below supports all formats:
first: output color strings;second: answer from given color strings;Correctness Proof
Let be the BFS distance from vertex .
For any edge , shortest-path distances satisfy:
The graph is bipartite, so adjacent vertices have opposite BFS parity. Therefore , and thus:
So every neighbor of a vertex lies either in layer or layer .
The agent colors each vertex by . For , some shortest path from to exists, so has at least one neighbor at distance .
A closer neighbor has color:
while a farther neighbor, if present, has color:
These colors are distinct modulo . Therefore, when two colors appear, one color corresponds exactly to closer neighbors and the other exactly to farther neighbors. The decoding table chooses the closer color in every possible pair.
When only one color appears, choosing any neighbor with that color is valid because at least one closer neighbor exists and all visible neighbors have the same color.
So every printed index points to a neighbor strictly closer to vertex .
Complexity
BFS costs per test case. Query handling costs linear time in the queried degree. Overall:
with memory.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
char colorOf(int dist) {
static const string colors = "rgb";
return colors[dist % 3];
}
int residue(char c) {
if (c == 'r') return 0;
if (c == 'g') return 1;
return 2;
}
char chooseColor(const string& s) {
bool seen[3] = {};
for (char c : s) seen[residue(c)] = true;
int cnt = seen[0] + seen[1] + seen[2];
if (cnt == 1) {
for (int i = 0; i < 3; i++) {
if (seen[i]) return colorOf(i);
}
}
for (int missing = 0; missing < 3; missing++) {
if (!seen[missing]) return colorOf((missing + 1) % 3);
}
return 'r';
}
void buildColors(int n, const vector<vector<int>>& adj, string& colors) {
vector<int> dist(n + 1, -1);
queue<int> q;
dist[1] = 0;
q.push(1);
while (!q.empty()) {
int v = q.front();
q.pop();
for (int u : adj[v]) {
if (dist[u] == -1) {
dist[u] = dist[v] + 1;
q.push(u);
}
}
}
colors.assign(n, 'r');
for (int i = 1; i <= n; i++) colors[i - 1] = colorOf(dist[i]);
}
void solveFirst() {
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1);
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
string colors;
buildColors(n, adj, colors);
cout << colors << '\n';
}
}
void solveSecond() {
int t;
cin >> t;
while (t--) {
int q;
cin >> q;
while (q--) {
int d;
string s;
cin >> d >> s;
char want = chooseColor(s);
for (int i = 0; i < d; i++) {
if (s[i] == want) {
cout << i + 1 << '\n';
break;
}
}
}
}
}
void solveHacked(int t) {
while (t--) {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1);
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
string colors;
buildColors(n, adj, colors);
int q;
cin >> q;
while (q--) {
int v;
cin >> v;
int d = (int)adj[v].size();
string seen;
seen.reserve(d);
for (int i = 0; i < d; i++) {
int p;
cin >> p;
int u = adj[v][p - 1];
seen.push_back(colors[u - 1]);
}
char want = chooseColor(seen);
for (int i = 0; i < d; i++) {
if (seen[i] == want) {
cout << i + 1 << '\n';
break;
}
}
}
}
}
int main() {
setIO();
string mode;
cin >> mode;
if (mode == "first") {
solveFirst();
} else if (mode == "second") {
solveSecond();
} else {
solveHacked(stoi(mode));
}
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
char colorOf(int dist) {
static const string colors = "rgb";
return colors[dist % 3];
}
int residue(char c) {
if (c == 'r') return 0;
if (c == 'g') return 1;
return 2;
}
char chooseColor(const string& s) {
bool seen[3] = {};
for (char c : s) seen[residue(c)] = true;
int cnt = seen[0] + seen[1] + seen[2];
if (cnt == 1) {
for (int i = 0; i < 3; i++) {
if (seen[i]) return colorOf(i);
}
}
for (int missing = 0; missing < 3; missing++) {
if (!seen[missing]) return colorOf((missing + 1) % 3);
}
return 'r';
}
void buildColors(int n, const vector<vector<int>>& adj, string& colors) {
vector<int> dist(n + 1, -1);
queue<int> q;
dist[1] = 0;
q.push(1);
while (!q.empty()) {
int v = q.front();
q.pop();
for (int u : adj[v]) {
if (dist[u] == -1) {
dist[u] = dist[v] + 1;
q.push(u);
}
}
}
colors.assign(n, 'r');
for (int i = 1; i <= n; i++) colors[i - 1] = colorOf(dist[i]);
}
void solveFirst() {
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1);
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
string colors;
buildColors(n, adj, colors);
cout << colors << '\n';
}
}
void solveSecond() {
int t;
cin >> t;
while (t--) {
int q;
cin >> q;
while (q--) {
int d;
string s;
cin >> d >> s;
char want = chooseColor(s);
for (int i = 0; i < d; i++) {
if (s[i] == want) {
cout << i + 1 << '\n';
break;
}
}
}
}
}
void solveHacked(int t) {
while (t--) {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n + 1);
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
string colors;
buildColors(n, adj, colors);
int q;
cin >> q;
while (q--) {
int v;
cin >> v;
int d = (int)adj[v].size();
string seen;
seen.reserve(d);
for (int i = 0; i < d; i++) {
int p;
cin >> p;
int u = adj[v][p - 1];
seen.push_back(colors[u - 1]);
}
char want = chooseColor(seen);
for (int i = 0; i < d; i++) {
if (seen[i] == want) {
cout << i + 1 << '\n';
break;
}
}
}
}
}
int main() {
setIO();
string mode;
cin >> mode;
if (mode == "first") {
solveFirst();
} else if (mode == "second") {
solveSecond();
} else {
solveHacked(stoi(mode));
}
}