树剖-边权挂点

原理简介

众所周知，树链剖分模板是用来维护点上信息的，但是也可通过对边的转化维护边的信息。
由于只有n-1条边，一般选择将所有边挂在dep[]更深的点上结果使根节点上无值，然后修改原来处理链的函数，核心代码：

if(x ^ y) {
    if(dep[x] > dep[y]) swap(x,y);
	ans += qsum(1,1,n,seg[son[x]],seg[y]);
} 

？？？为什么y一定在x的重儿子为根的子树上呢？我们考虑之前的while循环，设dep[top[x]] > dep[top[y]],所以每次都是把x - top[x]的支链求出答案，由假设得top[x]一定在以top[y]为根的子树内，但是top[x]管辖的链开头是自己，所以他并不是top[y]为根的子树内的重链。跳出循环时两点一定top[y]为根的子树的重链上。

例题

Housewife Wind

模板题

#include <iostream> 
#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>  
#define ll long long
#define rint register int
#define mid ((L + R) >> 1)
#define lson (x << 1)
#define rson (x << 1 | 1)
using namespace std;
template<typename xxx>inline void read(xxx &x) {
	int f = 1;char c = getchar();x = 0;
	for(;c ^ '-' && !isdigit(c);c = getchar());
	if(c == '-') f = -1,c = getchar();
	for(;isdigit(c);c = getchar()) x = (x<<3) + (x<<1) + (c ^ '0');
	x *= f;
} 
template<typename xxx>inline void print(xxx x) {
	if(x < 0) {
		putchar('-');
		x = -x;
	}
	if(x > 9) print(x / 10);
	putchar(x % 10 + '0');
}
const int maxn = 100010;
const int mod = 998244353;
const int inf = 0x7fffffff;
struct node{
	int a,b,id;ll c;
}g[maxn];
struct edge {
	int to,last;
	ll val;
}e[maxn<<1];
int head[maxn],tot;
inline void add(int from,int to,ll val) { 
	++tot;
	e[tot].to = to;
	e[tot].val = val;
	e[tot].last = head[from];
	head[from] = tot;
} 
int n,m,s;
int w[maxn],cnt;
int dad[maxn];
int dep[maxn];
int rev[maxn];
int seg[maxn];
int siz[maxn];
int son[maxn];
int top[maxn];
inline void ddfs1(int x,int da) {
	son[x] = 0;
	siz[x] = 1;dad[x] = da;
	dep[x] = dep[da] + 1;
	for(rint i = head[x];i;i = e[i].last) {
		if(e[i].to == da) continue;
		ddfs1(e[i].to,x);
		w[e[i].to] = g[e[i].val].c;
		g[e[i].val].id = e[i].to;
		siz[x] += siz[e[i].to];
		if(son[x] == 0 || siz[son[x]] < siz[e[i].to]) son[x] = e[i].to; 
	}
	return ;
}
inline void ddfs2(int x,int tp) {
	seg[x] = ++cnt;
	rev[cnt] = x;
	top[x] = tp;
	if(!son[x]) return ;
	ddfs2(son[x],tp);
	for(rint i = head[x];i;i = e[i].last) {
		if(e[i].to == dad[x] || e[i].to == son[x]) continue;
		ddfs2(e[i].to,e[i].to);
	}
	return ;
}
ll sum[maxn<<2];
inline void pushup(int x) {
	sum[x] = sum[lson] + sum[rson];
}
inline void build(int x,int L,int R) {
	sum[x] = 0;
	if(L == R) {
		sum[x] = w[rev[L]];
		return ; 
	}
	build(lson,L,mid);
	build(rson,mid + 1,R);
	pushup(x);
}
inline void update(int x,int L,int R,int pos,ll val) {
	if(L == R) {
		sum[x] = val;
		return;
	} 
	if(pos <= mid) update(lson,L,mid,pos,val);
	else update(rson,mid + 1,R,pos,val);
	pushup(x);
	return ;
}
inline ll qsum(int x,int L,int R,int l,int r) {
	if(l <= L && R <= r) return sum[x];
	ll ans = 0;
	if(l <= mid) ans += qsum(lson,L,mid,l,r);
	if(r >  mid) ans += qsum(rson,mid + 1,R,l,r);
	return ans;
}
inline ll q1(int x,int y) {
	ll ans = 0;
	while(top[x] ^ top[y]) {
		if(dep[top[x]] < dep[top[y]]) swap(x,y);
		ans += qsum(1,1,n,seg[top[x]],seg[x]);
		x = dad[top[x]];
	}
	if(x ^ y) {
		if(dep[x] > dep[y]) swap(x,y);
		ans += qsum(1,1,n,seg[son[x]],seg[y]);
	} 
	return ans;
}
int main() {
	while(~scanf("%d%d%d",&n,&m,&s)) {//小心可恶的多重数据
		for(rint i = 1;i < n; ++i) {
			read(g[i].a);read(g[i].b);read(g[i].c);
			add(g[i].a,g[i].b,i);
			add(g[i].b,g[i].a,i);
		}
		ddfs1(1,0);
		ddfs2(1,1);
		build(1,1,n);
		for(rint i = 1;i <= m; ++i) {
			ll opt,a,b;
			read(opt);
			if(opt) {
				read(a);read(b);
				update(1,1,n,seg[g[a].id],b);		
			} else {
				read(a);
				print(q1(a,s));
				putchar('
');
				s = a;
			}
		} 
		tot = 0;cnt = 0;
		memset(head,0,sizeof(head));
	}
	
	return 0;
}
/*

*/