SPOJ QTREE-Query on a tree

QTREE - Query on a tree

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONEOutput:13

题意：有一棵树，有两个操作，一种是改变某条边的权值，一种是查询点u到v之间的路径的最大边权

解题思路：树链剖分，边权给点权即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, x, y, z;int s[200009], nt[200009], e[200009], v[200009], cnt;int ct[200009], mx[200009], fa[200009], dep[200009];int top[200009], g[200009], G[200009], G1[200009], a[200009], ma[200009 << 2];char ch[10];void dfs(int x, int f){dep[x] = dep[f] + 1;fa[x] = f; ct[x] = 1; mx[x] = 0;for (int i = s[x]; ~i; i = nt[i]){if (e[i] == f) continue;dfs(e[i], x);ct[x] += ct[e[i]];if (ct[e[i]] > ct[mx[x]]) mx[x] = e[i];}}void Dfs(int x, int t){top[x] = !t ? x : top[fa[x]];g[x] = ++cnt;if (mx[x]) Dfs(mx[x], 1);for (int i = s[x]; ~i; i = nt[i]){if (e[i] == fa[x]) continue;if (e[i] == mx[x]) { G[g[e[i]]] = v[i]; G1[v[i]] = g[e[i]]; continue; }Dfs(e[i], 0);G[g[e[i]]] = v[i];G1[v[i]] = g[e[i]];}}void build(int k, int l, int r){if (l == r) { ma[k] = a[G[l]]; return; }int mid = (l + r) >> 1;build(k << 1, l, mid);build(k << 1 | 1, mid + 1, r);ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}void update(int k, int l, int r, int p, int val){if (l == r) { ma[k] = val; return; }int mid = (l + r) >> 1;if (p <= mid) update(k << 1, l, mid, p, val);else update(k << 1 | 1, mid + 1, r, p, val);ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}int query(int k, int l, int r, int ll, int rr){if (l >= ll&&r <= rr) return ma[k];int ans = 0;int mid = (l + r) >> 1;if (ll <= mid) ans = max(ans, query(k << 1, l, mid, ll, rr));if (rr > mid) ans = max(ans, query(k << 1 | 1, mid + 1, r, ll, rr));return ans;}int getans(int x, int y){int ans = 0;while (top[x] != top[y]){if (dep[top[x]] < dep[top[y]]) swap(x, y);ans = max(ans, query(1, 1, n, g[top[x]], g[x])); x = fa[top[x]];}if (dep[x] > dep[y]) swap(x, y);if (x == y) return ans;return max(ans, query(1, 1, n, g[x] + 1, g[y]));}int main(){int t;scanf("%d", &t);while (t--){scanf("%d", &n);memset(s, -1, sizeof s);dep[0] = ct[0] = cnt = 0;for (int i = 1; i < n; i++){scanf("%d%d%d", &x, &y, &a[i]);nt[cnt] = s[x], s[x] = cnt, v[cnt] = i, e[cnt++] = y;nt[cnt] = s[y], s[y] = cnt, v[cnt] = i, e[cnt++] = x;}dfs(1, 0);Dfs(1, cnt = 0);build(1, 1, n);while (scanf("%s", ch) && ch[0] != 'D'){scanf("%d%d", &x, &y);if (ch[0] == 'C') update(1, 1, n, G1[x], y);else printf("%d\n", getans(x, y));}}return 0;}