FZU 2082过路费 树链剖分

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

题意：Description
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。
Input
有多组样例，每组样例第一行输入两个正整数n,m(2 <= n<=50000，1<=m <= 50000),接下来n-1行，每行3个正整数a b c，(1 <= a,b <= n , a != b , 1 <= c <= 1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行，表示m个操作，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。

Output
对于每个询问，输出一行，表示最少要花的过路费。
Sample Input
2 3
1 2 1
1 1 2
0 1 2
1 2 1
Sample Output
1
2

思路：树链剖分，线段树维护区间和，入门题

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 30010;struct edge{    int to, next;}g[N*2];struct node{    int l, r, val, sum;}s[N*4];int dep[N], siz[N], son[N], top[N], fat[N], id[N], head[N];int d[N][3], val[N];int n, cnt, num;void add_edge(int v, int u){    g[cnt].to = u;    g[cnt].next = head[v];    head[v] = cnt++;}void dfs1(int v, int fa, int d){    dep[v] = d, fat[v] = fa, siz[v] = 1, son[v] = 0;    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fa)        {            dfs1(u, v, d + 1);            siz[v] += siz[u];            if(siz[son[v]] < siz[u])                son[v] = u;        }    }}void dfs2(int v, int tp){    top[v] = tp, id[v] = ++num;    if(son[v]) dfs2(son[v], tp);    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != son[v] && u != fat[v])            dfs2(u, u);    }}void push_up(int k){    s[k].val = s[k<<1].val + s[k<<1|1].val;}void build(int l, int r, int k){    s[k].l = l, s[k].r = r;    if(l == r)    {        s[k].val = val[l];        return;    }    int mid = (l + r) >> 1;    build(l, mid, k << 1);    build(mid + 1, r, k << 1|1);    push_up(k);}void update(int v, int c, int k){    if(s[k].l == s[k].r)    {        s[k].val = c; return;    }    int mid = (s[k].l + s[k].r) >> 1;    if(v <= mid) update(v, c, k << 1);    else update(v, c, k << 1|1);    push_up(k);}int seek(int l, int r, int k){    if(l <= s[k].l && s[k].r <= r)        return s[k].val;    int mid = (s[k].l + s[k].r) >> 1, ans = 0;    if(l <= mid) ans += seek(l, r, k << 1);    if(r > mid) ans += seek(l, r, k << 1 | 1);    return ans;}int query(int v, int u){    int t1 = top[v], t2 = top[u], ans = 0;    while(t1 != t2)    {        if(dep[t1] < dep[t2])            swap(t1, t2), swap(v, u);        ans += seek(id[t1], id[v], 1);        v = fat[t1];        t1 = top[v];    }    if(v == u) return ans;    if(dep[v] > dep[u]) swap(v, u);    return ans + seek(id[son[v]], id[u], 1);}int main(){    int m;    while(~scanf("%d%d", &n, &m))    {        memset(head, -1, sizeof head);        cnt = 0, num = 0;        for(int i = 1; i <= n - 1; i++)        {            scanf("%d%d%d", &d[i][0], &d[i][1], &d[i][2]);            add_edge(d[i][0], d[i][1]);            add_edge(d[i][1], d[i][0]);        }        dfs1(1, 0, 1);        dfs2(1, 1);        for(int i = 1; i <= n - 1; i++)        {            if(dep[d[i][0]] > dep[d[i][1]])                swap(d[i][0], d[i][1]);            val[id[d[i][1]]] = d[i][2];        }        build(1, num, 1);        int a, b, c;        for(int i = 0; i < m; i++)        {            scanf("%d%d%d", &a, &b, &c);            if(a == 0) update(id[d[b][1]], c, 1);            else printf("%d\n", query(b, c));        }    }    return 0;}