結果

問題 No.363 門松サイクル
ユーザー pekempey
提出日時 2016-04-18 15:51:48
言語 C++11(廃止可能性あり)
(gcc 13.3.0)
結果
AC  
実行時間 185 ms / 4,000 ms
コード長 2,713 bytes
コンパイル時間 1,395 ms
コンパイル使用メモリ 163,984 KB
実行使用メモリ 27,896 KB
最終ジャッジ日時 2024-10-04 13:37:55
合計ジャッジ時間 5,516 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 27
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:86:42: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   86 |         for (int i = 0; i < n; i++) scanf("%d", &a[i]);
      |                                     ~~~~~^~~~~~~~~~~~~
main.cpp:90:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   90 |                 scanf("%d %d", &x, &y);
      |                 ~~~~~^~~~~~~~~~~~~~~~~
main.cpp:103:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  103 |                 scanf("%d %d", &u, &v);
      |                 ~~~~~^~~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
bool isKadomatsu(long long x, long long y, long long z) {
if (x == y || y == z || x == z) return false;
if (x > y && y < z) return true;
if (x < y && y > z) return true;
return false;
}
int n;
int a[101010];
vector<int> g[101010];
int parent[24][101010];
int depth[101010];
bool kado[24][101010];
void dfs(int curr, int prev) {
parent[0][curr] = prev;
kado[0][curr] = true;
if (prev != -1 && parent[0][prev] != -1) {
int p = prev;
int pp = parent[0][prev];
kado[0][curr] = isKadomatsu(a[curr], a[p], a[pp]);
}
for (int next : g[curr]) if (next != prev) {
depth[next] = depth[curr] + 1;
dfs(next, curr);
}
}
void build() {
dfs(0, -1);
for (int i = 0; i < 23; i++) {
for (int j = 0; j < n; j++) {
if (parent[i][j] != -1) {
parent[i + 1][j] = parent[i][parent[i][j]];
kado[i + 1][j] = kado[i][j] && kado[i][parent[i][j]];
} else {
parent[i + 1][j] = -1;
kado[i + 1][j] = kado[i][j];
}
}
}
}
int ancestor(int u, int k) {
for (int i = 0; i < 24; i++) if (k & 1 << i) u = parent[i][u];
return u;
}
int lca(int u, int v) {
if (depth[u] < depth[v]) swap(u, v);
for (int i = 23; i >= 0; i--) {
if (depth[u] - depth[v] >= 1 << i) {
u = parent[i][u];
}
}
if (u == v) return u;
for (int i = 23; i >= 0; i--) {
if (parent[i][u] != parent[i][v]) {
u = parent[i][u];
v = parent[i][v];
}
}
return parent[0][u];
}
int query(int u, int d) {
if (d < 0) return true;
bool result = kado[0][u];
for (int i = 0; i < 24; i++) if (d & 1 << i) {
result &= kado[i][u];
u = parent[i][u];
result &= kado[0][u];
}
return result;
}
int main() {
cin >> n;
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
for (int i = 0; i < n - 1; i++) {
int x, y;
scanf("%d %d", &x, &y);
x--;
y--;
g[x].push_back(y);
g[y].push_back(x);
}
build();
int q;
cin >> q;
for (int i = 0; i < q; i++) {
int u, v;
scanf("%d %d", &u, &v);
u--;
v--;
int l = lca(u, v);
bool ans = true;
if (parent[0][u] == v || parent[0][v] == u) {
ans = false;
} else if (l != u && l != v) {
int p = ancestor(u, depth[u] - depth[l] - 1);
int q = ancestor(v, depth[v] - depth[l] - 1);
ans &= isKadomatsu(a[p], a[l], a[q]);
ans &= isKadomatsu(a[parent[0][u]], a[u], a[v]);
ans &= isKadomatsu(a[parent[0][v]], a[v], a[u]);
ans &= query(u, depth[u] - depth[l] - 2);
ans &= query(v, depth[v] - depth[l] - 2);
} else {
if (l == v) swap(u, v);
int q = ancestor(v, depth[v] - depth[l] - 1);
ans &= query(v, depth[v] - depth[u] - 2);
ans &= isKadomatsu(a[q], a[u], a[v]);
ans &= isKadomatsu(a[parent[0][v]], a[v], a[u]);
}
printf("%s\n", ans ? "YES" : "NO");
}
}
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