Submission #399258

#	TimeUTC-0	Username	Problem	Language	Result	Execution time	Memory
399258	2021-05-05 13:40:03	order999	Synchronization (JOI13_synchronization)	C++14	100 / 100	283 ms	22936 KiB

synchronization

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/*
  - Consider when we add an edge between nodes u and v that have a and b amounts of data, respectively, with c being the same
    - Both components of u and v will now have a + b - c amounts of data
  - c is just the amount of data shared between u and v the last time they were connected
  - We now need to support the following queries:
    - Add an edge
    - Delete an edge
    - Set values of all nodes in a connected component to x
    - Find the value of a node
  - Each connected component can be represented by the node that is closest to the root
  - Let's color / erase these nodes when we add / delete edges
    - Note that when we add an edge, one of its endpoints will become a representative
  - A connected component's representative can be found with binary search and path sum queries (using binary indexed tree)
  - For the set / find values queries, just process them on the representative
  - O(nlog^2n) since it's O(log^2n) for finding a representative
*/
#include <bits/stdc++.h>
using namespace std;
const int mxN=1e5;
int n, m, q, x[mxN-1], y[mxN-1], st[mxN], en[mxN], dt, ft[mxN+1], v[mxN], lc[mxN-1], anc[mxN][17], dj, ck;
vector<int> adj[mxN];
inline void upd(int i, int x) {
  for(++i; i<=n; i+=i&-i)
    ft[i]+=x;
}
inline int qry(int i) {
 
 
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• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 40 / 40