spoj375Query on a tree树链剖分

学了下树链剖分，这个链接讲的很详细http://blog.sina.com.cn/s/blog_6974c8b20100zc61.html

就是把一颗树划分轻重链，然后标号，对应到一颗线段树上。更新a到b的路径时，就把在下面的向上提 直到在一条重链上或者两点重合。

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <string.h>

using namespace std;

#define lson l,mid,rt<<1

#define rson mid+1,r,rt<<1|1





const int maxn = 111111;

struct Node

{

    int to;int val;int next;

}e[maxn];

int edge[maxn][3];

int len ;

int z; ;

int son[maxn];

int father[maxn];

int size[maxn];

int deep[maxn];

int pos[maxn],top[maxn];

int Max[maxn<<2];

int head[maxn];

void add(int from, int to, int val)

{

    e[len].to = to;

    e[len].val = val;

    e[len].next = head[from];

    head[from] = len++;

}





void init(int x)

{

    size[x] = 1; son[x] = 0;

    for (int i = head[x]; i != -1; i = e[i].next){

        int cc = e[i].to;

        if (cc == father[x]) continue;

        father[cc] = x; deep[cc] = deep[x] + 1;

        init(cc);

        if (size[cc] > size[son[x]]) son[x] = cc;

        size[x]+=size[cc];

    }

}



void dfs(int x, int tp)

{

    pos[x] = ++z; top[x] = tp;

    if (son[x]) dfs(son[x], tp);

    for (int i = head[x]; i != -1; i = e[i].next){

        int cc = e[i].to;

        if (cc == son[x] || cc == father[x]) continue;

        dfs(cc, cc);

    }

}



void up(int rt)

{

    Max[rt] = max(Max[rt << 1], Max[rt << 1 | 1]);

}

void update(int pos, int key, int l, int r, int rt)

{

    if (l == r){

        Max[rt] = key; return;

    }

    int mid = (l + r) >> 1;

    if (pos <= mid) update(pos, key, lson);

    else update(pos, key, rson);

    up(rt);

}



int ask(int L, int R, int l, int r, int rt)

{

    if (L <= l&&r <= R) return Max[rt];

    int mid = (l + r) >> 1;

    int ans = -1;

    if (L <= mid) ans = max(ans, ask(L, R, lson));

    if (R > mid) ans = max(ans, ask(L, R, rson));

    return ans;

}



int gao(int x, int y)

{

    int ans = -1;

    int f1 = top[x]; int f2 = top[y];

    while (f1 != f2){

        if (deep[f1] < deep[f2]){

            swap(f1, f2);

            swap(x, y);

        }

        ans = max(ans, ask(pos[f1], pos[x], 1, z, 1));

        x = father[f1]; f1 = top[x];

    }

    if (x == y) return ans;

    if (deep[x]>deep[y]) swap(x, y);

    ans = max(ans ,ask(pos[son[x]],pos[y],1,z,1));

    return ans;

}





int main()

{

    char str[100];

    int Icase;

    int n;

    cin >> Icase;

    while (Icase--){

        cin >> n;

        memset(size,0,sizeof(size));

        memset(Max,0,sizeof(Max));

        z = 0; len = 0;

        memset(head, -1, sizeof(head));

        for (int i = 1; i < n; i++){

            scanf("%d%d%d", &edge[i][0], &edge[i][1], &edge[i][2]);

            add(edge[i][0], edge[i][1], edge[i][2]);

            add(edge[i][1], edge[i][0], edge[i][2]);

        }

        init(1);

        dfs(1, 1);

        for (int i = 1; i < n ; i++){

            int a = edge[i][0]; int b = edge[i][1]; int c = edge[i][2];

            if (deep[a]>deep[b]) swap(edge[i][0], edge[i][1]);

            update(pos[edge[i][1]], c, 1, z, 1);

        }

        while (scanf("%s", str) != EOF){

            if (str[0] == 'D') break;

            int a; int b;

            scanf("%d%d", &a, &b);

            if (str[0] == 'C'){

                update(pos[edge[a][1]], b, 1, z, 1);

            }

            else {

                printf("%d\n", gao(a, b));

            }

        }

    }

    return 0;

}