# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
841105 | danikoynov | 봉쇄 시간 (IOI23_closing) | C++17 | 48 ms | 10032 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "closing.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3e3 + 10;
vector < pair < int, ll > > adj[maxn];
ll dist[2][maxn];
int n;
void bfs(int s, ll arr[])
{
queue < int > q;
for (int i = 0; i < n; i ++)
arr[i] = -1;
arr[s] = 0;
q.push(s);
while(!q.empty())
{
int v = q.front();
q.pop();
for (pair < int, ll > nb : adj[v])
{
if (arr[nb.first] == -1)
{
arr[nb.first] = arr[v] + nb.second;
q.push(nb.first);
}
}
}
}
const ll inf = 1e18;
ll value[maxn];
ll dp[maxn][maxn], sub[maxn], temp[maxn];
void dfs(int v, int par)
{
sub[v] = 1;
for (pair < int, ll > nb : adj[v])
{
if (nb.first == par)
continue;
dfs(nb.first, v);
}
dp[v][0] = 0;
dp[v][1] = max((ll)(0), dist[1][v] - value[v]);
for (pair < int, ll > nb : adj[v])
{
if (nb.first == par)
continue;
int u = nb.first;
for (int i = 0; i <= sub[u] + sub[v]; i ++)
temp[i] = inf;
for (int i = 0; i <= sub[v]; i ++)
for (int j = 0; j <= sub[u]; j ++)
{
if (i == 0 && j > 0)
continue;
temp[i + j] = min(temp[i + j], dp[v][i] + dp[u][j]);
}
for (int i = 0; i <= sub[v] + sub[u]; i ++)
dp[v][i] = temp[i];
sub[v] += sub[u];
}
}
int max_score(int N, int X, int Y, long long K,
vector<int> U, vector<int> V, vector<int> W)
{
for (int i = 0; i < N; i ++)
{
dist[0][i] = dist[1][i] = 0;
adj[i].clear();
value[i] = 0;
}
n = N;
for (int i = 0; i < N - 1; i ++)
{
adj[U[i]].push_back({V[i], W[i]});
adj[V[i]].push_back({U[i], W[i]});
}
bfs(X, dist[0]);
bfs(Y, dist[1]);
vector < pair < ll, int > > ord;
ll left_k = K;
for (int i = 0; i < N; i ++)
{
ord.push_back({dist[0][i], i});
}
sort(ord.begin(), ord.end());
///for (int i = 0; i < ord.size(); i ++)
/// cout << "here " << ord[i].first << " " << ord[i].second << endl;
ll ans = 0;
for (int i = 0; i < ord.size(); i ++)
{
left_k = left_k - ord[i].first;
if (left_k < 0)
break;
value[ord[i].second] = ord[i].first;
dfs(Y, -1);
/**cout << "step " << i + 1 << " " << left_k << endl;
for (int j = 0; j <= N; j ++)
cout << dp[Y][j] << " ";
cout << endl;*/
ll cur = 0;
int j = 0;
while(j <= n && dp[Y][j] <= left_k)
j ++;
j --;
cur = i + 1 + j;
if (cur > ans)
ans = cur;
}
return ans;
}
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