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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 |
---|---|---|---|---|---|---|---|---|---|
#300937 | #7103. Red Black Tree | kasrata | RE | 2ms | 10236kb | C++20 | 3.1kb | 2024-01-09 02:16:16 | 2024-01-09 02:16:17 |
Judging History
answer
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
#define F first
#define S second
#define pb push_back
#define ll long long
#define pil pair<int, ll>
#define pii pair<int, int>
#define vi vector<int>
const int N = 1e5 + 7, LG = 18;;
int n, m, q, s, leaf, root, h[N], st[N], en[N], red[N], spar[N][LG];
ll hw[N];
vector<pii> ad[N];
pil dis[N];
void rm() {
s = 0;
for (int i = 0; i < n; i++) {
ad[i].clear();
st[i] = en[i] = h[i] = hw[i] = 0;
red[i] = false;
for (int j = 0; j < LG; j++)
spar[i][j] = 0;
dis[i] = {0, 0};
}
}
void input() {
cin >> n >> m >> q;
for (int i = 0, r; i < m; i++)
cin >> r, r--, red[r] = true;
for (int i = 0, u, v, w; i < n - 1; i++) {
cin >> u >> v >> w, u--, v--;
ad[u].pb({v, w}), ad[v].pb({u, w});
}
}
void dfs(int v, int par, int who, ll len) {
dis[v] = {who, len};
st[v] = s++;
for (auto [u, w]: ad[v])
if (u != par) {
spar[u][0] = v;
hw[u] = hw[v] + w;
h[u] = h[v] + 1;
red[u]? dfs(u, v, u, 0): dfs(u, v, who, len + w);
}
en[v] = s;
}
bool isPar(int u, int v) {
return st[v] >= st[u] && en[u] >= en[v];
}
int lca(int u, int v) {
if (h[u] < h[v])
swap(u, v);
for (int j = LG - 1; j >= 0; j--)
if ((1 << j) <= h[u] - h[v])
u = spar[u][j];
if (u == v)
return v;
for (int j = LG - 1; j >= 0; j--)
if (spar[u][j] != spar[v][j])
u = spar[u][j], v = spar[v][j];
return spar[u][0];
}
bool cmp1(int u, int v) {
return -dis[u].S < -dis[v].S;
}
bool cmp2(int u, int v) {
return -h[lca(u, leaf)] < -h[lca(v, leaf)];
}
bool cmp3(int u, int v) {
return h[u] < h[v];
}
vi uniq(vi v) {
vi res;
sort(v.begin(), v.end());
res.pb(v[0]);
for (int i = 1; i < (int) v.size(); i++)
if (v[i] != v[i - 1])
res.pb(v[i]);
return res;
}
int cal(vi vec, int k) {
if (k == 1)
return 0;
sort(vec.begin(), vec.end(), cmp1);
if (dis[vec[0]].S == dis[vec[1]].S && dis[vec[0]].F != dis[vec[1]].F)
return dis[vec[0]].S;
vi vc;
multiset<ll> s;
vi row;
ll mx_out = 0;
leaf = vec[0], root = dis[leaf].F;
for (int v: vec)
if (dis[v].F == root) {
int u = lca(v, leaf);
vc.pb(v), s.insert(-dis[v].S), row.pb(u);
}
else
mx_out = max(mx_out, dis[v].S);
row = uniq(row);
sort(vc.begin(), vc.end(), cmp2);
sort(row.begin(), row.end(), cmp3);
ll mx1 = 0, mn = 1e18;
for (int v: vc) {
int u = row.back();
if (lca(v, leaf) != u) {
ll w = s.empty()? 0: -(*s.begin());
mn = min(mn, max({mx1, mx_out, w}));
row.pop_back();
mx1 += hw[u] - hw[row.back()];
}
s.erase(s.find(-dis[v].S));
mx1 = max(mx1, hw[v] - hw[row.back()]);
}
return min(mn, max(mx_out, mx1));
}
int main() {
ios:: sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T;
for (cin >> T; T; T--) {
input(), dfs(0, -1, 0, 0);
for (int i = 0; i < n; i++)
for (int j = 1; j < LG; j++)
spar[i][j] = spar[spar[i][j - 1]][j - 1];
while (q--) {
int k;
cin >> k;
vi v;
for (int i = 0, u; i < k; i++)
cin >> u, u--, v.pb(u);
cout << cal(v, k) << '\n';
}
rm();
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 2ms
memory: 10236kb
input:
2 12 2 4 1 9 1 2 1 2 3 4 3 4 3 3 5 2 2 6 2 6 7 1 6 8 2 2 9 5 9 10 2 9 11 3 1 12 10 3 3 7 8 4 4 5 7 8 4 7 8 10 11 3 4 5 12 3 2 3 1 2 1 2 1 1 3 1 1 1 2 1 2 3 1 2 3
output:
4 5 3 8 0 0 0
result:
ok 7 lines
Test #2:
score: -100
Runtime Error
input:
522 26 1 3 1 1 4 276455 18 6 49344056 18 25 58172365 19 9 12014251 2 1 15079181 17 1 50011746 8 9 2413085 23 24 23767115 22 2 26151339 26 21 50183935 17 14 16892041 9 26 53389093 1 20 62299200 24 18 56114328 11 2 50160143 6 26 14430542 16 7 32574577 3 16 59227555 3 15 8795685 4 12 5801074 5 20 57457...
output:
148616264 148616264 0 319801028 319801028 255904892 317070839 1265145897 1265145897 1072765445 667742619 455103436 285643094 285643094 285643094 317919339 0 894940447 691421476 605409472 479058444 371688030 303203698 493383271 919185207 910180170 919185207 121535083 181713164 181713164 181713164 181...