Submission #474246

#	TimeUTC-0	Username	Problem	Language	Result	Execution time	Memory
474246	2021-09-17 13:47:07	hoanghq2004	Magic Tree (CEOI19_magictree)	C++14	100 / 100	141 ms	32680 KiB

magictree

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/*
    Central European Olympiad in Informatics 2019 (Day 2)
    Using dynamic programming: dp[u][t] is the maximum of fruits can be harvested at time point earlier than or equal t
    Formula: dp[u][t] = sum(dp[v][t]) (if t < t[u])
                      = max(sum(dp[v][t]), sum(dp[v][t[u]]) + w[u]) otherwise
    So easy to find that: dp[u][t] is an non-decreasing array with t = 0...k and the number of distinct values will smaller than or equal the number of vertex in the subtree rooted u + 1
and the values only can be increase if t = t[v]
    Optimize:
        - map <int, long long> opt[u] contains all values of dp[u][i] - dp[u][i - 1]
        - merge small to large so the complexity is O(n * log^2(n))
        - for each subtree rooted u, dp[u][t[u]] += w[u], so opt[u][t[u]] += w and the values after t[u] have to be increase that: dp[u][t] = max(dp[u][t], w[u]).
          here, define cur = w[u] - dp[u][t] with t = t[u]...
          while opt[u][t] < w[u], dp[u][t] is smaller than w[u], so we should assign dp[u][t] = w[u]
          subtract opt[u][t] from cur and assign opt[u][t] = 0 (erase from map)
          in other case, dp[u][t] is larger than w[u] then all values from t shouldn't be changed, subtract w[u] from opt[u][t]
*/
#include <bits/stdc++.h>
using namespace std;
int n, m, k;
vector <int> e[100010];
int w[100010], t[100010];
map <int, long long> opt[100010];
long long res;
void dfs(int u) {
    for (auto v: e[u]) {
 
 
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