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: 1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE Output: 1 3
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 ", 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