POJ 2763 Housewife Wind （樹鍊剖分+邊權化點權）

題意：給定一棵含n個結點的生成樹，共有q次操作，分為兩種：

0 c：求從位置s到c的距離，然後s變成c

1 a b：把第a條邊的權值變為b

題解：樹鍊剖分

0操作就是樹上區間查詢。

由于該題是邊權且為生成樹，樹鍊剖分隻能解決點權，那我們化成點權即可。即對于邊<u, v>，将離根節點遠的那個點的權值賦為邊權。

這樣在進行樹剖的時候，減去lca的點權即可。

這樣1操作就是單點更新了。

#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;
/*------------準備階段--------------*/
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);
    }
}
/*--------------樹狀數組--------------*/
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;
}
/*------------樹鍊剖分-------------*/
int getSum(int x, int y) {  //查詢x到y路徑上所有結點和
    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]);  //減去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;
}