Submission #299507

#	TimeUTC-0	Username	Problem	Language	Result	Execution time	Memory
299507	2020-09-15 05:14:19	shivensinha4	Putovanje (COCI20_putovanje)	C++17	110 / 110	211 ms	28664 KiB

putovanje

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/*
COCI 2020 Putovanje
- Essentially want to find the number of times we traverse each edge
- We use the fact that A->B = A->LCA(A, B)->B
    - A path A->B where B is an ancestor of A passes through the "parent"
      edge of C iff B is an ancestor of C and C is an ancestor of A
    - This means we can mark A with 1 and B with -1
    - So the subtree sum is the number of paths passing through an edge!
- DFS to get DFS order and use a Fenwick tree to do stuff
- Complexity: O(N log N)
*/
#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
int n;
vector<int> graph[200001];
ll bit[200001];
int tin[200001], tout[200001], timer = 0;
int anc[200001][20];
void dfs(int node = 1, int parent = 0) {
    anc[node][0] = parent;
    FOR(i, 1, 20) anc[node][i] = anc[anc[node][i - 1]][i - 1];
    tin[node] = ++timer;
    for (int i : graph[node]) if (i != parent) dfs(i, node);
    tout[node] = timer;
 
 
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• Subtask 1: 20 / 20
• Subtask 2: 25 / 25
• Subtask 3: 65 / 65