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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#671693 | #5088. Two Choreographies | KowerKoint# | WA | 1ms | 5672kb | C++23 | 2.5kb | 2024-10-24 14:05:33 | 2024-10-24 14:05:33 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int N;
int A[200000], B[200000];
vector<int> e[100000];
int doubling[100000][20];
int depth[100000];
set<pair<int, int>> used;
void calc(int i, int v) {
depth[i] = depth[v] + 1;
doubling[i][0] = v;
for (int j = 1; j < 20; j++) {
if (doubling[i][j - 1] == -1) {
doubling[i][j] = -1;
} else {
doubling[i][j] = doubling[doubling[i][j - 1]][j - 1];
}
}
used.insert({min(v, i), max(v, i)});
}
void dfs(int v) {
for (auto i : e[v]) {
if (doubling[i][0] != -2) continue;
calc(i, v);
dfs(i);
}
}
int lca(int a, int b) {
if (depth[a] < depth[b]) swap(a, b);
while (depth[a] > depth[b]) {
int j = 0;
int d = 1;
while (d * 2 <= depth[a] - depth[b]) {
j++;
d <<= 1;
}
a = doubling[a][j];
}
while (a != b) {
int j = 0;
while (doubling[a][j + 1] != doubling[b][j + 1]) j++;
a = doubling[a][j];
b = doubling[b][j];
}
return a;
}
int dist(int a, int b) {
return depth[a] + depth[b] - 2 * depth[lca(a, b)];
}
void print_path(int a, int b) {
int l = lca(a, b);
while (a != l) {
cout << a + 1 << " ";
a = doubling[a][0];
}
cout << l + 1;
vector<int> ans;
while (b != l) {
ans.push_back(b);
b = doubling[b][0];
}
reverse(ans.begin(), ans.end());
for (auto i : ans) cout << " " << i + 1;
}
int main() {
cin >> N;
for (int i = 0; i < 2 * N - 3; i++) {
cin >> A[i] >> B[i];
A[i]--;
B[i]--;
if (A[i] > B[i]) swap(A[i], B[i]);
e[A[i]].push_back(B[i]);
e[B[i]].push_back(A[i]);
}
int start_i = 0;
for (int i = 0; i < N; i++) {
doubling[i][0] = -2;
if (e[i].size() >= 3) start_i = i;
}
for (int j = 0; j < 20; j++) doubling[start_i][j] = -1;
calc(e[start_i][0], start_i);
calc(e[start_i][1], start_i);
calc(e[start_i][2], start_i);
dfs(e[start_i][0]);
dfs(e[start_i][1]);
dfs(e[start_i][2]);
dfs(start_i);
vector<int> inv_dist(N, -1);
for (int i = 0; i < 2 * N - 3; i++) {
if (used.contains({A[i], B[i]})) continue;
int d = dist(A[i], B[i]);
if (inv_dist[d] == -1) {
inv_dist[d] = i;
} else {
cout << d + 1 << endl;
// cerr << A[i] << " " << B[i] << " " << A[inv_dist[d]] << " " << B[inv_dist[d]] << endl;
print_path(A[i], B[i]);
cout << endl;
print_path(A[inv_dist[d]], B[inv_dist[d]]);
cout << endl;
break;
}
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3620kb
input:
4 1 2 1 3 1 4 2 3 2 4
output:
3 1 2 4 1 2 3
result:
ok
Test #2:
score: 0
Accepted
time: 1ms
memory: 5668kb
input:
5 1 2 1 3 1 4 1 5 2 3 2 5 3 4
output:
3 1 3 4 1 3 2
result:
ok
Test #3:
score: 0
Accepted
time: 1ms
memory: 5672kb
input:
7 1 2 3 4 5 6 5 2 3 1 6 1 4 2 4 5 2 6 3 6 1 5
output:
3 2 6 5 1 6 2
result:
ok
Test #4:
score: -100
Wrong Answer
time: 1ms
memory: 5636kb
input:
40 1 16 1 40 2 4 2 16 2 36 3 25 3 38 4 1 4 13 5 11 5 27 6 4 7 5 7 11 8 10 8 14 8 24 9 34 10 20 12 35 13 2 14 10 14 20 15 18 15 28 15 31 16 6 16 13 17 5 17 11 17 27 18 9 19 1 19 4 19 16 20 24 21 12 21 33 21 35 22 38 23 12 23 21 25 28 25 31 25 34 25 38 26 14 26 20 27 7 27 11 28 3 28 31 29 16 29 19 30 ...
output:
1 1 -1 40 1 -1 16
result:
wrong answer Integer 1 violates the range [3, 40]