SPOJ-375 QTREE

QTREE - Query on a tree

#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

#include <bits/stdc++.h>using namespace std;#define maxn 10004vector<vector<int> >g(maxn);struct edge{int from, to, val;}a[maxn];int c[maxn << 2], sz[maxn], id[maxn], dep[maxn], son[maxn], top[maxn],  pre[maxn], val[maxn], tot;void dfs(int x, int fa, int d){sz[x] = 1;pre[x] = fa;dep[x] = d;son[x] = 0;int cur;for(int i = 0; i < g[x].size(); ++i){cur = g[x][i];if(cur == fa) continue;dfs(cur, x, d + 1);sz[x] += sz[cur];if(sz[son[x]] < sz[cur]){son[x] = cur;}}}void dfs1(int x, int tp){id[x] = ++tot;top[x] = tp;if(son[x]) dfs1(son[x], tp);int cur;for(int i = 0; i < g[x].size(); ++i){cur = g[x][i];if(cur == pre[x] || cur == son[x]) continue;dfs1(cur, cur);}}void build(int o, int l, int r){if(l == r){c[o] = val[l];return;}int mid = l + r >> 1;build(o << 1, l, mid);build(o << 1 | 1, mid + 1, r);c[o] = max(c[o << 1], c[o << 1 | 1]);}int query(int o, int l, int r, int L, int R){if(l >= L & r <= R){return c[o];}int mid = l + r >> 1, ans = -1e9;if(mid >= L) ans = max(ans, query(o << 1, l, mid, L, R));if(mid < R) ans = max(ans, query(o << 1 | 1, mid + 1, r, L, R));return ans;}int getmx(int x, int y){int tp1 = top[x], tp2 = top[y];int ans = -1e9;while(tp1 != tp2){if(dep[tp1] < dep[tp2]){swap(tp1, tp2);swap(x, y);}ans = max(ans, query(1, 1, tot, id[tp1], id[x]));x = pre[tp1];tp1 = top[x];}if(x != y){if(dep[x] < dep[y]) swap(x, y);ans = max(ans, query(1, 1, tot, id[son[y]], id[x]));}return ans;}void add(int o, int l, int r, int pos, int v){if(l == r){c[o] = v;return;}int mid = l + r >> 1;if(pos <= mid) add(o << 1, l, mid, pos, v);else add(o << 1 | 1, mid + 1, r, pos, v);c[o] = max(c[o << 1], c[o << 1 | 1]);}int main(){int T, n, u, v, x;scanf("%d", &T);while(T--){scanf("%d", &n);for(int i = 1; i <= n; ++i){g[i].clear();}for(int i = 1; i <= n; ++i){scanf("%d %d %d", &u, &v, &x);g[u].push_back(v);g[v].push_back(u);a[i].from = u;a[i].to = v;a[i].val = x;}tot = 0;dfs(1, 0, 1);dfs1(1, 1);for(int i = 1; i <= n; ++i){if(dep[a[i].from] < dep[a[i].to]){swap(a[i].from, a[i].to);}val[id[a[i].from]] = a[i].val;}build(1, 1, tot);char s[10];while(scanf("%s", s) != EOF){if(s[0] == 'D'){break;}if(s[0] == 'Q'){scanf("%d %d", &u, &v);printf("%d\n", getmx(u, v));}if(s[0] == 'C'){scanf("%d %d", &u, &v);a[u].val = v;add(1, 1, tot, id[a[u].from], v);}}}}/*题意：一棵树，1e4个点，每条边有边权，然后多次操作，每次操作要么修改某条边的边权，要么查询两点之间路径上边权最大值。思路：树链剖分的模板题了。唯一不同的是这次权值是在边上，这样我们把每一条边的边权转移记录到深度较大的点上，因为每个点向上只有一条边，向下可能有多条边。这样我们就还是对点操作了。注意最后tp1 == tp2时，可能x != y,这样我们还需要查询x和y之间的边，因为我们的记录方式是把边值转移到深度较大的点上，所以y下面的那条边记录在son[y]上。为什么是son[y]而不是别的儿子呢？如果x和y在lca的两边，那么tp1 == tp2时，x一定等于y。如果x或y是lca就有可能会发生tp1 == tp2时x != y的情况，tp1 == tp2说明最后x,y的top值相同，说明它们在一条重链上，所以是son[y]。*/