POJ 2763 Housewife Wind （树链剖分+边权化点权）

题意：给定一棵含n个结点的生成树，共有q次操作，分为两种：
0 c：求从位置s到c的距离，然后s变成c
1 a b：把第a条边的权值变为b

题解：树链剖分
0操作就是树上区间查询。

由于该题是边权且为生成树，树链剖分只能解决点权，那我们化成点权即可。即对于边<u, v>，将离根节点远的那个点的权值赋为边权。

这样在进行树剖的时候，减去lca的点权即可。

这样1操作就是单点更新了。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
#define pii pair<int, int>
using namespace std;
const int MAXN = 1e5 + 5;
int n, q, s, x[MAXN], y[MAXN], w[MAXN];
int sum[MAXN << 2], add[MAXN << 2];
int head[MAXN << 1], pre[MAXN], id[MAXN];
int siz[MAXN], son[MAXN], fa[MAXN];
int dep[MAXN], top[MAXN];
ll delta[MAXN];
struct edge {
    
    int to;
    int next;
}e[MAXN << 1];
int tot, sign;
/*------------准备阶段--------------*/
void init() {
    
    memset(head, -1, sizeof(head));
    memset(id, 0, sizeof(id));
    memset(siz, 0, sizeof(siz));
    memset(son, 0, sizeof(son));
    memset(fa, 0, sizeof(fa));
    memset(dep, 0, sizeof(dep));
    memset(top, 0, sizeof(top));
    memset(delta, 0, sizeof(delta));
    tot = sign = 0;
}
void addedge(int u, int v) {
    
    e[tot].to = v, e[tot].next = head[u], head[u] = tot++;
}
/*--------------dfs-----------------*/
void dfs1(int u, int f) {
        //1 0
    dep[u] = dep[f] + 1;
    fa[u] = f;
    siz[u] = 1;
    for (int i = head[u]; i != -1; i = e[i].next) {
    
        int to = e[i].to;
        if (to == f) continue;
        dfs1(to, u);
        siz[u] += siz[to];
        if (siz[to] > siz[son[u]]) son[u] = to;
    }
}
void dfs2(int u, int Top) {
      //1 1
    id[u] = ++sign;
    pre[sign] = u;
    top[u] = Top;
    if (son[u]) dfs2(son[u], Top);
    for (int i = head[u]; i != -1; i = e[i].next) {
    
        int to = e[i].to;
        if (to == son[u] || to == fa[u]) continue;
        dfs2(to, to);
    }
}
/*--------------树状数组--------------*/
int tree[MAXN];
void update(int x, int num) {
    
    for (; x <= n; x += x & -x)
        tree[x] += num;;
}
int su(int x) {
    
    int answer = 0;
    for (; x > 0; x -= x & -x)
        answer += tree[x];
    return answer;
}
/*------------树链剖分-------------*/
int getSum(int x, int y) {
      //查询x到y路径上所有结点和
    ll res = 0;
    while (top[x] != top[y]) {
    
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        res += su(id[x]) - su(id[top[x]] - 1);
        x = fa[top[x]];
    }
    if (id[x] > id[y]) swap(x, y);
    res += su(id[y]) - su(id[x]);  //减去lca
    return res;
}

int main() {
    
    init();
	scanf("%d%d%d", &n, &q, &s);
	for (int i = 1; i < n; i++) {
    
		scanf("%d%d%d", &x[i], &y[i], &w[i]);
        addedge(x[i], y[i]);
        addedge(y[i], x[i]);
	}
    dfs1(1, 0);
    dfs2(1, 1);
    for (int i = 1; i < n; i++) {
    
        if (fa[x[i]] == y[i]) swap(x[i], y[i]);
        update(id[y[i]], w[i]);
    }
    int op, a, b;
	while (q--) {
    
		scanf("%d", &op);
		if (op == 0) {
    
			scanf("%d", &a);
            printf("%d\n", getSum(s, a));
            s = a;
		}
		else {
    
			scanf("%d%d", &a, &b);
            update(id[y[a]], b - w[a]);
            w[a] = b;
		}
	}
	return 0;
}