FZU 2082 过路费

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2082

题意：

有n座城市，由n-1条路相连通，使得任意两座城市之间可达。每条路有过路费，要交过路费才能通过。每条路的过路费经常会更新，现问你，当前情况下，从城市a到城市b最少要花多少过路费。

操作有两种：

一. 0 a b，表示更新第a条路的过路费为b，1 <= a <= n-1 ；

二. 1 a b ， 表示询问a到b最少要花多少过路费。

思路：

权值在边上的树链剖分+线段树单点更新+线段树成段询问

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define maxn 50010
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
int n, q, s;
struct Node
{
    int to;
    long long val;
    Node(int t, long long v)
    {
        to = t; val = v;
    }
    Node(){}
};
vector <Node> mp[maxn];
struct Edge
{
    int from, to;
    long long val;
}e[maxn];
int pos;
int siz[maxn], dep[maxn], top[maxn], fa[maxn], son[maxn], w[maxn], fw[maxn];
long long cnt[maxn<<2];
void PushUp(int rt)
{
    cnt[rt] = cnt[rt<<1] + cnt[rt<<1|1];
}
void build(int l, int r, int rt)
{
    cnt[rt] = 0;
    if(l == r) return;
    int m = (l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int p, long long val, int l, int r, int rt)
{
    if(l == r)
    {
        cnt[rt] = val; return;
    }
    int m = (l+r)>>1;
    if(p <= m) update(p, val, lson);
    else update(p, val, rson);
    PushUp(rt);
}
long long query(int L, int R, int l, int r, int rt)
{
    if(L <= l && R >= r)
    {
        return cnt[rt];
    }
    long long ret = 0;
    int m = (l+r)>>1;
    if(L <= m) ret += query(L, R, lson);
    if(R > m) ret += query(L, R, rson);
    return ret;
}
int dfs1(int u, int pre, int deep)
{
    siz[u] = 1; fa[u] = pre; dep[u] = deep;
    int mmax = 0;
    for(int i = 0; i < mp[u].size(); i++)
    {
        if(mp[u][i].to != pre)
        {
            int temp = dfs1(mp[u][i].to, u, deep+1);
            siz[u] += temp;
            if(son[u] == -1 || temp >= mmax) son[u] = mp[u][i].to;        
        }
    }
    return siz[u];
}
void dfs2(int u, int val)
{
    top[u] = val;
    if(son[u] != -1)
    {
        w[u] = pos++;
        fw[w[u]] = u;
        dfs2(son[u], val);
    }
    else if(son[u] == -1)
    {
        w[u] = pos++;
        fw[w[u]] = u;
        return;
    }
    for(int i = 0; i < mp[u].size(); i++)
    {
        if(mp[u][i].to != fa[u] && mp[u][i].to != son[u]) dfs2(mp[u][i].to, mp[u][i].to);
    }
}
long long find(int u, int v)
{
    int f1 = top[u], f2 = top[v];
    long long temp = 0;
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])
        {
            swap(f1, f2);
            swap(u, v);
        }
        temp += query(w[f1], w[u], 1, pos-1, 1);
        u = fa[f1]; f1 = top[u];
    }
    if(u == v) return temp;
    if(dep[u] > dep[v]) swap(u, v);
    temp += query(w[son[u]], w[v], 1, pos-1, 1);
    return temp;
}
int main() 
{
  // freopen("in.txt", "r", stdin);
   //freopen("out.txt", "w", stdout);
    while(~scanf("%d%d", &n, &q))
    {
        for(int i = 1; i <= n; i++) mp[i].clear();
        pos = 0; memset(son, -1, sizeof(son));
        
        for(int i = 1; i <= n-1; i++)
        {
            scanf("%d%d%lld", &e[i].from, &e[i].to, &e[i].val);
            mp[e[i].from].push_back(Node(e[i].to, e[i].val));
            mp[e[i].to].push_back(Node(e[i].from, e[i].val));
        }
        dfs1(1, -1, 1);
        dfs2(1, 1);
        build(1, pos-1, 1);
        
        for(int i = 1; i <= n-1; i++)
        {
            if(dep[e[i].from] > dep[e[i].to])
            {
                swap(e[i].from, e[i].to);
            }
            update(w[e[i].to], e[i].val, 1, pos-1, 1);
        }
        while(q--)
        {
            int op; scanf("%d", &op);
            if(op == 1)
            {
                //int u; scanf("%d", &u);
                int a, b; scanf("%d%d", &a, &b);
                printf("%lld
", find(a, b));
            }
            else if(op == 0)
            {
                int i;
                long long ww; scanf("%d%lld", &i, &ww);
                update(w[e[i].to], ww, 1, pos-1, 1);
            }
        }
    }
    return 0;
}