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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 |
---|---|---|---|---|---|---|---|---|---|
#695954 | #5439. Meet in the Middle | HRcohc | TL | 0ms | 0kb | C++14 | 4.1kb | 2024-10-31 21:09:19 | 2024-10-31 21:09:29 |
answer
#include <bits/stdc++.h>
#define N 100005
#define M 500005
#define ll long long
#define pb push_back
#define fst first
#define scd second
using namespace std;
inline int read(){
char ch = getchar(); int x = 0, f = 1;
while(!isdigit(ch)){if(ch == '-') f = -1; ch = getchar();}
while(isdigit(ch)){x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
return x * f;
}
int Ti, n, q, pos1, pos2;
int dfn[N], in[N], out[N], id[N], lg2[N << 1], st[20][N << 1];
ll d[N], dis[N], ans[M];
vector<pair<int, int>> a[N], b[N], qry[N];
inline int mn(int x, int y){return (id[x] < id[y])? x : y;}
void dfs(int u, int fa){
st[0][id[u] = ++pos1] = u;
for(auto tmp : b[u]){
int v = tmp.fst; ll w = tmp.scd;
if(v == fa) continue;
d[v] = d[u] + w;
dfs(v, u);
st[0][++pos1] = u;
}
}
void dfs2(int u, int fa){
in[u] = ++pos2, dfn[pos2] = u;
for(auto tmp : a[u]){
int v = tmp.fst; ll w = tmp.scd;
if(v == fa) continue;
dis[v] = dis[u] + w;
dfs2(v, u);
}
out[u] = pos2;
}
int lca(int u, int v){
if(u == v) return u;
if(id[u] > id[v]) swap(u, v);
int k = lg2[id[v] - id[u] + 1];
return mn(st[k][id[u]], st[k][id[v] - (1 << k) + 1]);
}
inline ll dist(int u, int v){return d[u] + d[v] - 2ll * d[lca(u, v)];}
inline ll D(int u, int v, ll d1, ll d2){return (u == v)? 0 : dist(u, v) + d1 + d2;}
struct Node{
int u, v;
ll d1, d2, tag;
friend Node operator + (const Node x, const Node y){
Node res = x;
if(D(y.u, y.v, y.d1, y.d2) > D(res.u, res.v, res.d1, res.d2)) res = y;
if(D(x.u, y.u, x.d1, y.d1) > D(res.u, res.v, res.d1, res.d2)) res = {x.u, y.u, x.d1, y.d1};
if(D(x.u, y.v, x.d1, y.d2) > D(res.u, res.v, res.d1, res.d2)) res = {x.u, y.v, x.d1, y.d2};
if(D(x.v, y.u, x.d2, y.d1) > D(res.u, res.v, res.d1, res.d2)) res = {x.v, y.u, x.d2, y.d1};
if(D(x.v, y.v, x.d2, y.d2) > D(res.u, res.v, res.d1, res.d2)) res = {x.v, y.v, x.d2, y.d2};
return res;
}
};
struct Segment{
#define Lc (now << 1)
#define Rc (now << 1 | 1)
#define mid ((l + r) >> 1)
Node t[N << 2];
void pushup(int now){int k = t[now].tag; t[now] = t[Lc] + t[Rc]; t[now].tag = k;}
void update(int now, ll k){t[now].d1 += k, t[now].d2 += k, t[now].tag += k;}
void pushdown(int now){if(t[now].tag) update(Lc, t[now].tag), update(Rc, t[now].tag), t[now].tag = 0;}
void build(int now = 1, int l = 1, int r = n){
t[now].tag = 0;
if(l == r){t[now] = {dfn[l], dfn[l], dis[dfn[l]], dis[dfn[l]]}; return;}
build(Lc, l, mid), build(Rc, mid + 1, r); pushup(now);
}
void modify(int Lt, int Rt, ll k, int now = 1, int l = 1, int r = n){
if(Lt > r || Rt < l) return;
if(Lt <= l && r <= Rt){update(now, k); return;}
pushdown(now); modify(Lt, Rt, k, Lc, l, mid), modify(Lt, Rt, k, Rc, mid + 1, r); pushup(now);
}
}T;
void dfs3(int u, int fa){
Node cur = T.t[1];
for(auto tmp : qry[u]) ans[tmp.scd] = max(D(cur.u, tmp.fst, cur.d1, 0), D(cur.v, tmp.fst, cur.d2, 0));
for(auto tmp : a[u]){
int v = tmp.fst; ll w = tmp.scd;
if(v == fa) continue;
T.modify(1, n, w), T.modify(in[v], out[v], -2ll * w);
dfs3(v, u);
T.modify(1, n, -w), T.modify(in[v], out[v], 2ll * w);
}
}
void init(){
pos1 = pos2 = 0;
for(int i = 1; i <= n; i++){
dfn[i] = in[i] = out[i] = id[i] = d[i] = dis[i] = 0;
vector<pair<int, int>>().swap(a[i]), vector<pair<int, int>>().swap(b[i]), vector<pair<int, int>>().swap(qry[i]);
}
}
void solve(){
n = read();
init();
for(int i = 1; i < n; i++){
int u = read(), v = read(), w = read();
a[u].pb({v, w}), a[v].pb({u, w});
}
for(int i = 1; i < n; i++){
int u = read(), v = read(), w = read();
b[u].pb({v, w}), b[v].pb({u, w});
}
dfs(1, 0); dfs2(1, 0);
for(int i = 1; i < 20; i++){
for(int j = 1; j + (1 << i) - 1 <= pos1; j++) st[i][j] = mn(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
}
q = read();
for(int i = 1; i <= q; i++){
int x = read(), y = read();
qry[x].pb({y, i});
}
T.build(); dfs3(1, 0);
for(int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
}
int main(){
// freopen("move.in", "r", stdin);
// freopen("move.out", "w", stdout);
for(int i = 2; i < N << 1; i++) lg2[i] = lg2[i >> 1] + 1;
Ti = read();
while(Ti--) solve();
return 0;
}
詳細信息
Test #1:
score: 0
Time Limit Exceeded
input:
3 4 1 2 1 2 3 2 1 2 2 2 3 1 1 1 1 2 2 1 2 2