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Mo's on tree (Query on Path)
YessineJallouli edited this page Nov 17, 2021
·
2 revisions
#include <bits/stdc++.h>
#define ll long long
#define SaveTime ios_base::sync_with_stdio(false), cin.tie(0);
using namespace std;
struct query {
int L,R,id;
query() {
L = R = id = 0;
}
query(int _L, int _R, int _id) {
L = _L;
R = _R;
id = _id;
}
};
int block;
bool cmp(query q1, query q2) {
if (q1.L/block < q2.L/block)
return true;
if (q1.L/block == q2.L/block && q1.R < q2.R)
return true;
return false;
}
const int N = 1e5+7;
const int LOG = 18;
vector<vector<int>> tree(N);
int depth[N];
int tin[N];
int tout[N];
int up[N][LOG];
bool visited[N];
vector<int> tim;
void dfs(int node,int par = -1) {
tim.push_back(node);
tin[node] = tim.size()-1;
for (int ch : tree[node]) {
if (ch == par)
continue;
depth[ch] = depth[node] + 1;
up[ch][0] = node;
for (int j = 1; j < LOG; j++) {
up[ch][j] = up[up[ch][j-1]][j-1];
}
dfs(ch, node);
}
tim.push_back(node);
tout[node] = tim.size()-1;
}
int lca(int a, int b) {
if (depth[a] > depth[b])
swap(a,b);
int k = depth[b]-depth[a];
for (int j = LOG-1; j >= 0; j--) {
if (k & (1<<j))
b = up[b][j];
}
if (a == b)
return b;
for (int j = LOG-1; j >= 0; j--) {
if (up[a][j] != up[b][j]) {
a = up[a][j];
b = up[b][j];
}
}
return up[a][0];
}
void ins(int node) {
// use the array visited to check if you visited the node or not
}
void del(int node) {
}
int main() {
int n,q; cin >> n >> q;
block = 2*n/sqrt(q)+1;
for (int i = 0; i < n-1; i++) {
int a,b; cin >> a >> b;
tree[a].push_back(b);
tree[b].push_back(a);
}
depth[1] = 0;
for (int i = 0; i < LOG; i++) {
up[1][i] = 1;
}
dfs(1);
query qr[q];
for (int i = 0; i < q; i++) {
int a,b; cin >> a >> b;
int l = lca(a,b);
if (tin[a] < tin[b])
swap(a,b);
if (l == b)
qr[i] = query(tin[b], tin[a], i);
else
qr[i] = query(tout[b], tin[a], i);
}
sort(qr, qr+q, cmp);
int ans[q];
int l = 0, r = -1;
for (int i = 0; i < q; i++) {
auto [L,R,id] = qr[i];
while (r < R) {
r++;
ins(tim[r]);
}
while (r > R) {
del(tim[r]);
r--;
}
while (l < L) {
del(tim[l]);
l++;
}
while (l > L) {
l--;
ins(tim[l]);
}
int n1 = tim[L], n2 = tim[R];
int lc = lca(n1, n2);
/// check the LCA
}
for (int i = 0; i < q; i++) {
printf("%d\n", ans[i]);
}
}
Problems :
https://www.spoj.com/problems/COT2/en/