이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
typedef pair<int,int> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define maxn 100010
#define INF (ll)1e18
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;
int N,S,Q,E,I,R;
int A[maxn], B[maxn], W[maxn], x;
bool shop[maxn];
vpi adj[maxn];
int co = 0;
int st[maxn], en[maxn], depth[maxn];
void dfs(int x, int p){
st[x] = co++;
aFOR(i, adj[x]) if (i.f != p){
depth[i.f] = depth[x] + 1;
dfs(i.f,x);
}
en[x] = co - 1;
}
vpi queries[maxn];
int ans[maxn];
vi C[maxn];
int dist[maxn];
int32_t main(){
cin >> N >> S >> Q >> E;
FOR(i,1,N-1){
cin >> A[i] >> B[i] >> W[i];
adj[A[i]].pb(pi(B[i], W[i]));
adj[B[i]].pb(pi(A[i], W[i]));
}
FOR(i,1,S){
cin >> x; shop[x] = 1;
}
dfs(E,-1);
FOR(i,1,N-1){
if (depth[A[i]] > depth[B[i]]) swap(A[i], B[i]); // A is the parent
}
FOR(i,1,Q){
cin >> I >> R;
if (!(st[B[I]] <= st[R] && st[R] <= en[B[I]])) ans[i] = -2;
else{
queries[B[I]].pb(pi(R, i));
}
}
FOR(i,1,N) C[depth[i]].pb(i);
priority_queue <pi, vector <pi>, greater <pi> > pq, pq2;
mem(dist, -1);
DEC(i,N-1,0){
while (!pq2.empty()){
pq.push(pq2.top()); pq2.pop();
}
aFOR(j, C[i]){
if (shop[j]){
dist[j] = 0; pq.push(pi(0, j));
}
}
while (!pq.empty()){
pi cur = pq.top(); pq.pop();
if (cur.f != dist[cur.s]) continue;
if (depth[cur.s] == i-1){
pq2.push(cur); continue;
}
aFOR(j, adj[cur.s]) if (dist[j.f] == -1 || dist[j.f] > cur.f + j.s){
dist[j.f] = cur.f + j.s;
pq.push(pi(dist[j.f], j.f));
}
}
aFOR(j, C[i]){
aFOR(k, queries[j]){
ans[k.s] = dist[k.f];
}
}
}
FOR(i,1,Q){
if (ans[i] == -2) cout << "escaped\n";
else if (ans[i] == -1) cout << "oo\n";
else cout << ans[i] << '\n';
}
}
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