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1fd0050 | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 | #include <bits/stdc++.h>
using namespace std;
static vector<vector<int>> create_map(int N, int M, vector<int> A, vector<int> B) {
vector<vector<int>> g(N + 1);
vector<vector<char>> adj(N + 1, vector<char>(N + 1, 0));
for (int i = 0; i < M; i++) {
int u = A[i], v = B[i];
g[u].push_back(v);
g[v].push_back(u);
adj[u][v] = adj[v][u] = 1;
}
if (N == 1) {
int K = 3;
return vector<vector<int>>(K, vector<int>(K, 1));
}
// Build a spanning tree with BFS.
vector<int> parent(N + 1, 0);
vector<int> order;
order.reserve(N);
queue<int> q;
parent[1] = -1;
q.push(1);
while (!q.empty()) {
int u = q.front(); q.pop();
order.push_back(u);
for (int v : g[u]) {
if (parent[v] == 0) {
parent[v] = u;
q.push(v);
}
}
}
// Problem guarantees a valid map exists; for N>1 this implies connectivity.
// If not connected, still attempt a fallback by connecting components via any existing edge path (best effort).
if ((int)order.size() != N) {
// Best-effort: connect remaining nodes arbitrarily to 1 (may violate constraints if truly disconnected).
for (int v = 1; v <= N; v++) if (parent[v] == 0) parent[v] = 1;
}
vector<vector<int>> tree(N + 1);
for (int v = 2; v <= N; v++) {
int p = parent[v];
if (p > 0) {
tree[p].push_back(v);
tree[v].push_back(p);
}
}
// Euler tour on the tree (vertex sequence), consecutive vertices are tree-adjacent.
vector<int> seq;
seq.reserve(2 * N);
function<void(int,int)> dfs = [&](int u, int p) {
seq.push_back(u);
for (int v : tree[u]) {
if (v == p) continue;
dfs(v, u);
seq.push_back(u);
}
};
dfs(1, 0);
int L = (int)seq.size();
const int W = 3;
int K = W * L;
if (K > 240) {
// Shouldn't happen with N<=40 and this construction (K<=237). Fallback compress width.
// Reduce W to 2 (still gives an interior column only if W>=3, so keep safe by capping K and stretching rows).
// But constraints imply this won't be needed.
K = 240;
}
vector<vector<int>> C(K, vector<int>(K, 1));
for (int t = 0; t < L; t++) {
int color = seq[t];
for (int col = W * t; col < W * t + W && col < K; col++) {
for (int row = 0; row < K; row++) C[row][col] = color;
}
}
vector<vector<char>> realized(N + 1, vector<char>(N + 1, 0));
for (int t = 0; t + 1 < L; t++) {
int a = seq[t], b = seq[t + 1];
int u = min(a, b), v = max(a, b);
realized[u][v] = 1;
}
vector<vector<int>> stripes(N + 1);
stripes.assign(N + 1, {});
for (int t = 0; t < L; t++) stripes[seq[t]].push_back(t);
vector<int> nextStripeIdx(N + 1, 0);
vector<int> rowPtr(L, 1);
vector<vector<char>> usedRow(L, vector<char>(K, 0)); // usedRow[stripe][row]
auto find_row = [&](int stripe) -> int {
int start = rowPtr[stripe];
auto ok = [&](int r) -> bool {
if (r < 1 || r > K - 2) return false;
if (usedRow[stripe][r]) return false;
if (usedRow[stripe][r - 1]) return false;
if (usedRow[stripe][r + 1]) return false;
return true;
};
for (int pass = 0; pass < 2; pass++) {
for (int r = start; r <= K - 2; r += 2) {
if (ok(r)) {
rowPtr[stripe] = r + 2;
return r;
}
}
start = 1;
}
for (int r = 1; r <= K - 2; r++) {
if (ok(r)) {
rowPtr[stripe] = r + 1;
return r;
}
}
return 1; // should never happen
};
for (int i = 0; i < M; i++) {
int u = A[i], v = B[i];
if (realized[u][v]) continue;
// Place a single cell of v inside an interior column of a stripe of u.
if (stripes[u].empty()) continue; // should not happen
int idx = nextStripeIdx[u]++ % (int)stripes[u].size();
int t = stripes[u][idx];
int col = W * t + 1;
if (col <= 0 || col >= K - 1) col = min(max(W * t, 0), K - 1); // best-effort fallback
int r = find_row(t);
C[r][col] = v;
usedRow[t][r] = 1;
realized[u][v] = 1;
}
return C;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
if (!(cin >> T)) return 0;
for (int tc = 0; tc < T; tc++) {
int N, M;
cin >> N >> M;
vector<int> A(M), B(M);
for (int i = 0; i < M; i++) cin >> A[i] >> B[i];
auto C = create_map(N, M, A, B);
int K = (int)C.size();
cout << K << "\n";
for (int i = 0; i < K; i++) {
if (i) cout << ' ';
cout << K;
}
cout << "\n";
for (int i = 0; i < K; i++) {
for (int j = 0; j < K; j++) {
if (j) cout << ' ';
cout << C[i][j];
}
cout << "\n";
}
if (tc + 1 < T) cout << "\n";
}
return 0;
} |