spoj 375 树链剖分（裸模板）

看来以后树链剖分要用这个模板了。具体效果怎么样，还要看看，机场wifi太TM慢了。

寒假估计要是有史以来最辛苦的了。寒假疯狂AC万岁。

//树链剖分模板/*记siz[v]表示以v为根的子树的节点数，dep[v]表示v的深度(根深度为1)，top[v]表示v所在的链的顶端节点，fa[v]表示v的父亲，son[v]表示与v在同一重链上的v的儿子节点（姑且称为重儿子），w[v]表示v与其父亲节点的连边（姑且称为v的父边）在线段树中的位置。只要把这些东西求出来，就能用logn的时间完成原问题中的操作。重儿子：siz[u]为v的子节点中siz值最大的，那么u就是v的重儿子。轻儿子：v的其它子节点。重边：点v与其重儿子的连边。轻边：点v与其轻儿子的连边。重链：由重边连成的路径。轻链：轻边。剖分后的树有如下性质：性质1：如果(v,u)为轻边，则siz[u] * 2 < siz[v]；性质2：从根到某一点的路径上轻链、重链的个数都不大于logn。*///spoj 375 树链剖分#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#define maxn 100000using namespace std;struct Tree{int l, r, _max;}tree[maxn*4];struct Eage{int to, next;}eage[maxn*4];int son[maxn], fa[maxn], top[maxn], six[maxn],e[maxn],w[maxn],dep[maxn],si[maxn][4];//son:the son in the same hign chain;fa:the father of node;ro:the first of the chain;//w:the number of the chain between the fanther and it;si:used to remember the information of the eage.int n,cnt,z;void add(int a, int b, int c){eage[++cnt].to = b;eage[cnt].next = e[a];e[a] = cnt;}void build_tree(int a, int b)              //用来给每个边进行编号（重边要连续），并且确定每个点的祖先节点。{w[a] = ++z;top[a] = b;if (son[a] != 0)build_tree(son[a], b);for (int i = e[a];i != -1;i = eage[i].next)if (eage[i].to != son[a] && eage[i].to != fa[a])build_tree(eage[i].to, eage[i].to);}void dfs(int v){six[v] = 1;son[v] = 0;for (int i = e[v];i != -1;i = eage[i].next){if (eage[i].to == fa[v])continue;fa[eage[i].to] = v;dep[eage[i].to] = dep[v] + 1;dfs(eage[i].to);if (six[eage[i].to] > six[son[v]])son[v] = eage[i].to;six[v] += six[eage[i].to];    }}void update(int a, int l, int r, int m, int v)  //对树链进行编号后的线段树进行初始化{if (l == r){tree[a]._max = v;return;}int mid = (l + r) / 2;if (m <= mid)update(a * 2, l, mid, m, v);elseupdate(a * 2 + 1, mid + 1, r, m, v);tree[a]._max = max(tree[a * 2]._max, tree[a * 2 + 1]._max);}void init(){for (int i = 1;i <= maxn * 4;i++)tree[i]._max = 0;scanf("%d", &n);cnt = 0;z = 0;int root = (1 + n) / 2;dep[root] =fa[root]= 0;for (int i = 1;i < n;i++){scanf("%d%d%d", &si[i][0], &si[i][1], &si[i][2]);add(si[i][0], si[i][1], si[i][2]);add(si[i][1], si[i][0], si[i][2]);        }dfs(root);build_tree(root, root);for (int i = 1;i < n;i++){if (dep[si[i][0]]>dep[si[i][1]])swap(si[i][0], si[i][1]);update(1, 1, z, w[si[i][1]], si[i][2]);        }}int maxi(int a, int l, int r, int m, int v){int ans = 0;if ( m<=l&&r <= v)return tree[a]._max;int mid = (l + r) / 2;if (m <= mid)ans =max(ans, maxi(a * 2, l, mid, m, v));if (v > mid)ans =max(ans, maxi(a * 2 + 1, mid + 1, r, m, v));//printf("maxi %d %d %d %d %d ans=%d\n", a, l, r, m, v, ans);return ans;}int query(int a, int b){int f1 = top[a], f2 = top[b], tmp = 0;while (f1 != f2){if (dep[f1] < dep[f2]){swap(f1, f2);swap(a, b);}tmp = max(tmp, maxi(1, 1, z, w[f1], w[a]));a = fa[f1];f1 = top[a];}if (a == b)return tmp;if (dep[a] > dep[b])swap(a, b);return max(tmp, maxi(1, 1, z, w[son[a]], w[b]));}void work(){char ch[20];scanf("%s", ch);while (ch[0] != 'D'){int a, b;scanf("%d%d", &a, &b);if (ch[0] == 'Q')printf("%d\n", query(a, b));elseupdate(1, 1, z, w[si[a][1]], b);scanf("%s", ch);}}int main(){int T;//freopen("d:\\in.txt", "r", stdin);//freopen("d:\\inn.txt", "w", stdout);scanf("%d", &T);while (T){        memset(e, -1, sizeof(e));memset(son, 0, sizeof(son));memset(top, 0, sizeof(top));memset(six, 0, sizeof(six));init();work();T--;}return 0;}

徒手敲了一个下午，很好