LCA + 树状数组 + 树上RMQ

题目链接：http://poj.org/problem?id=2763

思路：首先求出树上dfs序列，并且标记树上每个节点开始遍历以及最后回溯遍历到的时间戳，由于需要修改树上的某两个节点之间的权值，如果parent[v] = u, 那么说明修改之后的v的子树到当前根的距离都会改变，由于遍历到v时有开始时间戳以及结束时间戳，那么处于这个区间所有节点都会影响到，于是我们可以通过数组数组来更新某个区间的值，只需令从区间起始点之后的每个值都增加一个改变量了，令区间中止点之后的每个值都减小一个改变量，这样我们就可以得到处于该区间的值的改变量。至于如何求树上两个节点的LCA,这里可以使用RMQ,O(n
* log(n))的预处理，O(1)的查询复杂度。\

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAX_N = (200000 + 10000);
struct Edge {
    int u, v, w, next;
    Edge () {}
    Edge (int _u, int _v, int _w, int _next) : u(_u), v(_v), w(_w), next(_next) {}
} EE[MAX_N << 2], edge[MAX_N << 2];

int NE, head[MAX_N];
void Init()
{
    NE = 0;
    memset(head, -1, sizeof(head));
}

void Insert(int u, int v, int w)
{
    edge[NE].u = u;
    edge[NE].v = v;
    edge[NE].w = w;
    edge[NE].next = head[u];
    head[u] = NE++;
}

int index, dfs_seq[MAX_N << 2]; //dfs序列
int First[MAX_N], End[MAX_N]; //First数组表示第一次遍历到的时间戳，End数组表示最后回溯时遍历到的时间戳
int parent[MAX_N], dep[MAX_N << 2]; //深度
int N, Q, st, cost[MAX_N];

void dfs(int u, int fa, int d, int c)
{
    parent[u] = fa;
    First[u] = index;
    dfs_seq[index] = u;
    dep[index++] = d;
    cost[u] = c;

    for (int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].v, w = edge[i].w;
        if (v == fa) continue;
        dfs(v, u, d + 1, cost[u] + w);
        dfs_seq[index] = u;
        dep[index++] = d;
    }

    End[u] = index;
}

int dp[MAX_N][40]; //dp[i][j]表示从时间戳i开始，长度为(1 << j)的区间中深度最小的点的时间戳
void Init_RMQ()
{
    for (int i = 0; i < index; ++i) dp[i][0] = i;
    for (int j = 1; (1 << j) < index; ++j) {
        for (int i = 0; i + (1 << j) - 1 < index; ++i) {
            dp[i][j] = (dep[ dp[i][j - 1] ] < dep[ dp[i + (1 << (j - 1))][j - 1] ] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1]);
        }
    }
 }

 int RMQ_Query(int l, int r)
 {
     int k = (int)(log(r * 1.0 - l + 1) / log(2.0));
     return (dep[ dp[l][k] ] < dep[ dp[r - (1 << k) + 1][k] ] ? dp[l][k] : dp[r - (1 << k) + 1][k]);
 }

 int LCA(int x, int y)
 {
     if (x > y) swap(x, y);
     return dfs_seq[RMQ_Query(x, y)];
 }

int C[MAX_N << 2];
int lowbit(int x)
{
    return x & (-x);
}
void update(int i, int val)
{
    while (i < index) {
        C[i] += val;
        i += lowbit(i);
    }
}

int getSum(int i)
{
    int sum = 0;
    while (i > 0) {
        sum += C[i];
        i -= lowbit(i);
    }
    return sum;
}

 int main()
 {
     while (~scanf("%d %d %d", &N, &Q, &st)) {
        Init();
        for (int i = 1; i < N; ++i) {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            Insert(u, v, w);
            Insert(v, u, w);
            EE[i] = Edge(u, v, w, -1);
        }

        index = 0;
        dfs(st, st, 0, 0);

        Init_RMQ();

        memset(C, 0, sizeof(C));
        for (int i = 1; i <= Q; ++i) {
            int opt;
            scanf("%d", &opt);
            if (opt == 0) {
                int ed, fa, x, y, z;
                scanf("%d", &ed);
                x = First[st], y = First[ed], z = First[fa = LCA(x, y)];
                printf("%d\n", getSum(x + 1) + getSum(y + 1) - 2 * getSum(z + 1) + cost[st] + cost[ed] - 2 * cost[fa]);
                st = ed;

            } else {
                int id, w;
                scanf("%d %d", &id, &w);
                int u = EE[id].u, v = EE[id].v, tmp = w - EE[id].w;
                EE[id].w = w;

                if (parent[u] == v) swap(u, v);
                update(First[v] + 1, tmp);
                update(End[v] + 1, -tmp);
            }
        }
     }
     return 0;
 }