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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 |
---|---|---|---|---|---|---|---|---|---|
#378701 | #8575. Three Person Tree Game | ucup-team2112# | TL | 0ms | 3636kb | C++20 | 4.3kb | 2024-04-06 14:16:18 | 2024-04-06 14:16:19 |
Judging History
answer
#include <bits/stdc++.h>
struct HLD {
int n;
std::vector<int> sz, top, dep, parent, in, out, seq;
std::vector<std::vector<int>> adj;
int dfs_clock;
HLD() {}
HLD(int n) {
init(n);
}
void init(int n) {
this->n = n;
sz.resize(n);
top.resize(n);
dep.resize(n);
parent.resize(n);
in.resize(n);
out.resize(n);
seq.resize(n);
dfs_clock = 0;
adj.assign(n, {});
}
void addEdge(int u, int v) {
adj[u].push_back(v);
adj[v].push_back(u);
}
void work(int root = 0) {
top[root] = root;
dep[root] = 0;
parent[root] = -1;
dfs1(root);
dfs2(root);
}
void dfs1(int u) {
if (parent[u] != -1) {
adj[u].erase(std::find(adj[u].begin(), adj[u].end(), parent[u]));
}
sz[u] = 1;
for (auto &v : adj[u]) {
parent[v] = u;
dep[v] = dep[u] + 1;
dfs1(v);
sz[u] += sz[v];
if (sz[v] > sz[adj[u][0]]) {
std::swap(v, adj[u][0]);
}
}
}
void dfs2(int u) {
in[u] = dfs_clock++;
seq[in[u]] = u;
for (auto v : adj[u]) {
top[v] = v == adj[u][0] ? top[u] : v;
dfs2(v);
}
out[u] = dfs_clock;
}
int lca(int u, int v) {
while (top[u] != top[v]) {
if (dep[top[u]] > dep[top[v]]) {
u = parent[top[u]];
} else {
v = parent[top[v]];
}
}
return dep[u] < dep[v] ? u : v;
}
int dist(int u, int v) {
return dep[u] + dep[v] - 2 * dep[lca(u, v)];
}
// jump from u to k-th ancester
int jump(int u, int k) {
if (dep[u] < k) {
return -1;
}
int d = dep[u] - k;
while (dep[top[u]] > d) {
u = parent[top[u]];
}
return seq[in[u] - dep[u] + d];
}
// check if u is ancester of v (include u == v)
bool isAncester(int u, int v) {
return in[u] <= in[v] && in[v] < out[u];
}
// root is u, find parent of v
int rootedParent(int u, int v) {
std::swap(u, v);
if (u == v) {
return u;
}
if (!isAncester(u, v)) {
return parent[u];
}
auto it = std::upper_bound(adj[u].begin(), adj[u].end(), v, [&](int x, int y) {
return in[x] < in[y];
}) - 1;
return *it;
}
// root is u, find size of subtree rooted at v
int rootedSize(int u, int v) {
if (u == v) {
return n;
}
if (!isAncester(v, u)) {
return sz[v];
}
return n - sz[rootedParent(u, v)];
}
// find the middle point of a, b, c
int rootedLca(int a, int b, int c) {
return lca(a, b) ^ lca(b, c) ^ lca(c, a);
}
};
void solve(){
int n;
std::cin >> n;
HLD hld(n);
int a, b, c;
std::cin >> a >> b >> c;
a--, b--, c--;
for (int i = 1; i < n; i++) {
int u, v;
std::cin >> u >> v;
hld.addEdge(u - 1, v - 1);
}
hld.work();
while(true) {
if (hld.dist(a, c) <= 1) {
std::cout << "A\n";
return;
}
int na = hld.rootedParent(c, a);
if (hld.dist(na, b) <= 1) {
na = a;
}
if (hld.dist(na, b) <= 1) {
std::cout << "B\n";
return;
}
int nb = hld.rootedParent(na, b);
if (hld.dist(nb, c) <= 1) {
nb = b;
}
if (hld.dist(nb, c) <= 1) {
std::cout << "C\n";
return;
}
int nc = hld.rootedParent(nb, c);
if (hld.dist(nc, a) <= 1) {
nc = c;
}
if (na == a && nb == b && nc == c) {
std::cout << "DRAW\n";
return;
}
}
}
int main(){
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
int T;
std::cin >> T;
while(T--) {
solve();
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3636kb
input:
2 3 1 2 3 2 1 3 1 4 1 2 3 1 4 2 4 3 4
output:
A DRAW
result:
ok 2 lines
Test #2:
score: -100
Time Limit Exceeded
input:
10000 20 2 12 1 16 15 3 2 16 17 14 13 11 12 9 8 10 9 18 17 6 5 18 19 13 12 5 4 7 6 20 19 14 15 3 4 11 10 1 2 8 7 20 12 13 1 18 13 12 11 19 15 17 16 10 14 4 2 15 11 6 5 3 2 4 13 20 8 11 9 3 7 14 16 5 8 5 4 9 6 10 3 1 2 17 2 17 15 9 10 5 4 9 8 2 11 6 7 8 7 13 4 2 3 6 15 5 6 17 8 2 1 3 4 16 7 14 5 3 12...