首页 > 代码库 > [spoj 375]QTREE - Query on a tree[树链剖分]

[spoj 375]QTREE - Query on a tree[树链剖分]

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

树链剖分
  1 #include <cstdio>  2 #include <algorithm>  3 #include <iostream>  4 #include <string.h>  5 using namespace std;  6 const int maxn = 10010;  7 struct Tedge  8 { int b, next; } e[maxn * 2];  9 int tree[maxn]; 10 int zzz, n, z, edge, root, a, b, c; 11 int d[maxn][3]; 12 int first[maxn], dep[maxn], w[maxn], fa[maxn], top[maxn], son[maxn], siz[maxn]; 13 char ch[10]; 14  15 void insert(int a, int b, int c){ 16     e[++edge].b = b; 17     e[edge].next = first[a]; 18     first[a] = edge; 19 } 20  21 void dfs(int v){ 22     siz[v] = 1; son[v] = 0; 23     for (int i = first[v]; i > 0; i = e[i].next) 24         if (e[i].b != fa[v]){ 25             fa[e[i].b] = v; 26             dep[e[i].b] = dep[v]+1; 27             dfs(e[i].b); 28             if (siz[e[i].b] > siz[son[v]]) son[v] = e[i].b; 29             siz[v] += siz[e[i].b]; 30         } 31 } 32  33 void build_tree(int v, int tp){ 34     w[v] = ++ z; top[v] = tp; 35     if (son[v] != 0) build_tree(son[v], top[v]); 36     for (int i = first[v]; i > 0; i = e[i].next) 37         if (e[i].b != son[v] && e[i].b != fa[v]) 38             build_tree(e[i].b, e[i].b); 39 } 40  41 void update(int root, int lo, int hi, int loc, int x){ 42     if (loc > hi || lo > loc) return; 43     if (lo == hi) 44     { tree[root] = x; return; } 45     int mid = (lo + hi) / 2, ls = root * 2, rs = ls + 1; 46     update(ls, lo, mid, loc, x); 47     update(rs, mid+1, hi, loc, x); 48     tree[root] = max(tree[ls], tree[rs]); 49 } 50  51 int maxi(int root, int lo, int hi, int l, int r){ 52     if (l > hi || r < lo) return 0; 53     if (l <= lo && hi <= r) return tree[root]; 54     int mid = (lo + hi) / 2, ls = root * 2, rs = ls + 1; 55     return max(maxi(ls, lo, mid, l, r), maxi(rs, mid+1, hi, l, r)); 56 } 57  58 inline int find(int va, int vb){ 59     int f1 = top[va], f2 = top[vb], tmp = 0; 60     while (f1 != f2){ 61         if (dep[f1] < dep[f2]) 62         { swap(f1, f2); swap(va, vb); } 63         tmp = max(tmp, maxi(1, 1, z, w[f1], w[va])); 64         va = fa[f1]; f1 = top[va]; 65     } 66     if (va == vb) return tmp; 67     if (dep[va] > dep[vb]) swap(va, vb); 68     return max(tmp, maxi(1, 1, z, w[son[va]], w[vb]));  // 69 } 70  71 void init(){ 72     scanf("%d", &n); 73     root = (n + 1) / 2; 74     fa[root] = z = dep[root] = edge = 0; 75     memset(siz, 0, sizeof(siz)); 76     memset(first, 0, sizeof(first)); 77     memset(tree, 0, sizeof(tree)); 78     for (int i = 1; i < n; i++){ 79         scanf("%d%d%d", &a, &b, &c); 80         d[i][0] = a; d[i][1] = b; d[i][2] = c; 81         insert(a, b, c); 82         insert(b, a, c); 83     } 84     dfs(root); 85     build_tree(root, root);    // 86     for (int i = 1; i < n; i++){ 87         if (dep[d[i][0]] > dep[d[i][1]]) swap(d[i][0], d[i][1]); 88         update(1, 1, z, w[d[i][1]], d[i][2]); 89     } 90 } 91  92 inline void read(){ 93     ch[0] =  ; 94     while (ch[0] < C || ch[0] > Q) scanf("%s", &ch); 95 } 96  97 void work() 98 { 99     for (read(); ch[0] != D; read()){100         scanf("%d%d", &a, &b);101         if (ch[0] == Q) printf("%d\n", find(a, b));102         else update(1, 1, z, w[d[a][1]], b);103     }104 }105 106 int main(){107     for (scanf("%d", &zzz); zzz > 0; zzz--){108         init();109         work();110     }111     return 0;112 }

你让我变得苍白

却依然天真的相信花儿会再次的盛开

[spoj 375]QTREE - Query on a tree[树链剖分]