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hld.cpp
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hld.cpp
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#include <bits/stdc++.h>
#include <vector>
using namespace std;
#define root 0
#define N 10100
#define LN 14
vector <int> adj[N], costs[N], indexx[N];
int baseArray[N], ptr;
int chainNo, chainInd[N], chainHead[N], posInBase[N];
int depth[N], pa[LN][N], otherEnd[N], subsize[N];
int st[N*6];
/*
* make_tree:
* Used to construct the segment tree.
It uses the baseArray for construction
*/
// Initiall 1,0,n
void make_tree(int cur, int s, int e) {
// all nodes
if(s == e-1) {
st[cur] = baseArray[s];
return;
}
int c1 = (cur<<1), c2 = c1 | 1, m = (s+e)>>1;
make_tree(2*cur, s, m);
make_tree(2*cur+1, m, e);
st[cur] = st[c1] > st[c2] ? st[c1] : st[c2];
}
/*
* update_tree:
* Point update. Update a single element of the segment tree.
*/
void update_tree(int cur, int s, int e, int x, int val) {
if(s > x || e <= x) return;
if(s == x && s == e-1) {
st[cur] = val;
return;
}
int c1 = (cur<<1), c2 = c1 | 1, m = (s+e)>>1;
//cout << c1 << " " << c2 << " " << cur << endl;
update_tree(c1, s, m, x, val);
update_tree(c2, m, e, x, val);
st[cur] = st[c1] > st[c2] ? st[c1] : st[c2];
}
/*
* query_tree:
* Given S and E, it will return the maximum value in the range [S,E)
*/
int query_tree(int cur, int s, int e, int S, int E) {
if(s >= E || e <= S) {
return -1;
}
if(s >= S && e <= E) {
return st[cur];
}
int c1 = (cur<<1), c2 = c1 | 1, m = (s+e)>>1;
return max(query_tree(c1, s, m, S, E),query_tree(c2, m, e, S, E));
}
/*
* query_up:
* It takes two nodes u and v, condition is that v is an ancestor of u
* We query the chain in which u is present till chain head, then move to next chain up
* We do that way till u and v are in the same chain, we query for that part of chain and break
*/
int query_up(int u, int v) {
if(u == v) return 0; // Trivial
int uchain, vchain = chainInd[v], ans = -1;
// uchain and vchain are chain numbers of u and v
while(1) {
uchain = chainInd[u];
if(uchain == vchain) {
// Both u and v are in the same chain, so we need to query from u to v, update answer and break.
// We break because we came from u up till v, we are done
if(u==v) break;
ans = max(ans,query_tree(1, 0, ptr, posInBase[v]+1, posInBase[u]+1));
// Above is call to segment tree query function
//if(qt[1] > ans) ans = qt[1]; // Update answer
break;
}
ans = max(ans,query_tree(1, 0, ptr, posInBase[chainHead[uchain]], posInBase[u]+1));
// Above is call to segment tree query function. We do from chainHead of u till u. That is the whole chain from
// start till head. We then update the answer
//if(qt[1] > ans) ans = qt[1];
u = chainHead[uchain]; // move u to u's chainHead
u = pa[0][u]; //Then move to its parent, that means we changed chains
}
return ans;
}
/*
* LCA:
* Takes two nodes u, v and returns Lowest Common Ancestor of u, v
*/
int LCA(int u, int v) {
if(depth[u] < depth[v]) swap(u,v);
int diff = depth[u] - depth[v];
for(int i=0; i<LN; i++) if( (diff>>i)&1 ) u = pa[i][u];
if(u == v) return u;
for(int i=LN-1; i>=0; i--) if(pa[i][u] != pa[i][v]) {
u = pa[i][u];
v = pa[i][v];
}
return pa[0][u];
}
void query(int u, int v) {
/*
* We have a query from u to v, we break it into two queries, u to LCA(u,v) and LCA(u,v) to v
*/
int lca = LCA(u, v);
int ans = query_up(u, lca); // One part of path
int temp = query_up(v, lca); // another part of path
if(temp > ans) ans = temp; // take the maximum of both paths
printf("%d\n", ans);
}
/*
* change:
* We just need to find its position in segment tree and update it
*/
void change(int i, int val) {
int u = otherEnd[i];
update_tree(1, 0, ptr, posInBase[u], val);
}
/*
* Actual HL-Decomposition part
* Initially all entries of chainHead[] are set to -1.
* So when ever a new chain is started, chain head is correctly assigned.
* As we add a new node to chain, we will note its position in the baseArray.
* In the first for loop we find the child node which has maximum sub-tree size.
* The following if condition is failed for leaf nodes.
* When the if condition passes, we expand the chain to special child.
* In the second for loop we recursively call the function on all normal nodes.
* chainNo++ ensures that we are creating a new chain for each normal child.
*/
void HLD(int curNode, int cost, int prev) {
if(chainHead[chainNo] == -1) {
chainHead[chainNo] = curNode;
}
chainInd[curNode] = chainNo;
// Position of this node in
// baseArray which we will use in Segtree
posInBase[curNode] = ptr;
baseArray[ptr++] = cost;
int sc = -1, ncost;
// Loop to find special child
for(int i=0; i<adj[curNode].size(); i++) if(adj[curNode][i] != prev) {
if(sc == -1 || subsize[sc] < subsize[adj[curNode][i]]) {
sc = adj[curNode][i];
ncost = costs[curNode][i];
}
}
// If not leaf node
if(sc != -1) {
// Expand the chain to heavier nodes
HLD(sc, ncost, curNode);
}
// Expand for lighter nodes
for(int i=0; i<adj[curNode].size(); i++)
if(adj[curNode][i] != prev)
{
if(sc != adj[curNode][i])
{
// New chains a each normal node
chainNo++;
HLD(adj[curNode][i], costs[curNode][i], curNode);
}
}
}
/*
* dfs used to set parent of a node, depth of a node, subtree size of a node
*/
void dfs(int cur, int prev, int _depth=0) {
pa[0][cur] = prev;
depth[cur] = _depth;
subsize[cur] = 1;
for(int i=0; i<adj[cur].size(); i++)
if(adj[cur][i] != prev) {
otherEnd[indexx[cur][i]] = adj[cur][i];
dfs(adj[cur][i], cur, _depth+1);
subsize[cur] += subsize[adj[cur][i]];
}
}
int main() {
#ifndef ONLINE_JUDGE
// for getting input from input.txt
freopen("input.txt", "r", stdin);
// for writing output to output.txt
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while(t--) {
ptr = 0;
int n;
cin >> n;
// Cleaning step, new test case
for(int i=0; i<n; i++) {
adj[i].clear();
costs[i].clear();
indexx[i].clear();
chainHead[i] = -1;
for(int j=0; j<LN; j++) pa[j][i] = -1;
}
for(int i=1; i<n; i++) {
int u, v, c;
cin >> u >> v >> c;
u--; v--;
adj[u].push_back(v);
costs[u].push_back(c);
indexx[u].push_back(i-1);
adj[v].push_back(u);
costs[v].push_back(c);
indexx[v].push_back(i-1);
}
chainNo = 0;
dfs(root, -1); // We set up subsize, depth and parent for each node
HLD(root, -1, -1); // We decomposed the tree and created baseArray
//cout << ptr << endl;
make_tree(1, 0, n); // We use baseArray and construct the needed segment tree
// Below Dynamic programming code is for LCA.
for(int i=1; i<LN; i++)
for(int j=0; j<n; j++)
if(pa[i-1][j] != -1)
pa[i][j] = pa[i-1][pa[i-1][j]];
while(1) {
string s;
cin >> s;
if(s[0]=='D') {
break;
}
int a, b;
cin >> a >> b;
if(s[0]=='Q') {
query(a-1, b-1);
} else {
change(a-1, b);
}
}
}
}