树链剖分的学习2 例题

首先肯定是要先选好模板，并且把模板理解透彻

这个树链剖分的模板题 洛谷出的很好 https://www.luogu.org/problem/P3384

这个题目涉及了树链剖分的各种各样的操作

模板

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <stack>
#include <map>
#include <string>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 2e5 + 10;
 
int f[maxn];//f 保存u的父亲节点
int dep[maxn];//dep保存节点u 的深度
int siz[maxn];//siz保存以u为根的子节点的个数
int son[maxn];//son 保存u的重儿子
int rk[maxn];//rk当前dfs序在树中所对应的节点
int top[maxn];// top保存当前结点所在链的顶端结点
int id[maxn];//dfs的执行顺序
 
int a[maxn];
int mod, n;
int sum[maxn * 4], lazy[maxn * 4];
//------------------线段树部分---------------//
void push_up(int id)
{
    sum[id] = (sum[id << 1] + sum[id << 1 | 1]) % mod;
    // printf("sum[%d]=%d sum[%d]=%d
", id << 1, sum[id << 1], id << 1 | 1, sum[id << 1 | 1]);
    // printf("sum[%d]=%d
", id, sum[id]);
}
 
void build(int id,int l,int r)
{
    lazy[id] = 0;
    if(l==r)
    {
        sum[id] = a[rk[l]] % mod;
        // printf("id=%d sum=%d
", id, sum[id]);
        return;
    }
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    push_up(id);
}
 
void push_down(int id,int len1,int len2)
{
    if (lazy[id] == 0) return;
    sum[id << 1] += lazy[id] % mod * len1%mod;
    sum[id << 1] %= mod;
 
    lazy[id << 1] += lazy[id] % mod;
    lazy[id << 1] %= mod;
 
    sum[id << 1 | 1] += lazy[id] % mod * len2%mod;
    sum[id << 1 | 1] %= mod;
 
    lazy[id << 1 | 1] += lazy[id] % mod;
    lazy[id << 1 | 1] %= mod;
 
    lazy[id] = 0;
}
 
void update(int id,int l,int r,int x,int y,int val)
{
    // printf("id=%d l=%d r=%d x=%d y=%d val=%d
", id, l, r, x, y, val);
    if(x<=l&&y>=r)
    {
        // printf("id=%d sum=%d
", id, sum[id]);
        sum[id] += val * (r - l + 1) % mod;
        sum[id] %= mod;
        lazy[id] += val;
        lazy[id] %= mod;
        // printf("%d
", sum[id]);
        return;
    }
    int mid = (l + r) >> 1;
    push_down(id, mid - l + 1, r - mid);
    if (x <= mid) update(id << 1, l, mid, x, y, val);
    if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val);
    push_up(id);
}
 
int query(int id,int l,int r,int x,int y)
{
    if (x <= l && y >= r) return sum[id];
    int mid = (l + r) >> 1, ans = 0;
    push_down(id, mid - l + 1, r - mid);
    if (x <= mid) ans = (ans + query(id << 1, l, mid, x, y)) % mod;
    if (y > mid) ans = (ans + query(id << 1 | 1, mid + 1, r, x, y)) % mod;
    return ans;
}
 
 
 
//------------------------树链剖分-------------------//
// int f[maxn];//f 保存u的父亲节点
// int dep[maxn];//dep保存节点u 的深度
// int siz[maxn];//siz保存以u为根的子节点的个数
// int son[maxn];//son 保存u的重儿子
// int rk[maxn];//rk当前dfs序在树中所对应的节点
// int top[maxn];// top保存当前结点所在链的顶端结点
// int id[maxn];//dfs的执行顺序
struct node
{
    int v, nxt;
    node(int v=0,int nxt=0):v(v),nxt(nxt){}
}ex[maxn];
int head[maxn], cnt = 0, tot;
void init()
{
    cnt = 0, tot = 0;
    memset(son, 0, sizeof(son));
    memset(head, -1, sizeof(head));
}
void add(int u,int v)
{
    ex[cnt] = node(v, head[u]);
    head[u] = cnt++;
    ex[cnt] = node(u, head[v]);
    head[v] = cnt++;
}
 
 
void dfs1(int u,int fa,int depth)
{
    f[u] = fa; dep[u] = depth; siz[u] = 1;
    for(int i=head[u];i!=-1;i=ex[i].nxt)
    {
        int v = ex[i].v;
        if (v == fa) continue;
        dfs1(v, u, depth + 1);
        siz[u] += siz[v];
        if (siz[v] > siz[son[u]]) son[u] = v;
    }
}
 
void dfs2(int u,int t)
{
    top[u] = t;
    id[u] = ++tot;//标记dfs序
    rk[tot] = u;//序号tot对应的结点u
    if (!son[u]) return;
    dfs2(son[u], t);
    /*我们选择优先进入重儿子来保证一条重链上各个节点dfs序连续，
    一个点和它的重儿子处于同一条重链，所以重儿子所在重链的顶端还是t*/
    for(int i=head[u];i!=-1;i=ex[i].nxt)
    {
        int v = ex[i].v;
        if (v != son[u] && v != f[u]) dfs2(v, v);//一个点位于轻链底端，那么它的top必然是它本身
    }
}
 
void update2(int x,int y,int z)//修改x到y路径的值
{
    while(top[x]!=top[y])//不在同一条链上
    {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);//x为深度大的链
        update(1, 1, n, id[top[x]], id[x], z);//x为深度大的链
        x = f[top[x]];//深度大的向上跳
    }
    if (dep[x] > dep[y]) swap(x, y); //这里x和y在同一条链
    update(1, 1, n, id[x], id[y], z); //x和y这条链的更新
}
 
int query2(int x,int y)
{
    int ret = 0;
    while(top[x]!=top[y])
    {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ret = (ret + query(1, 1, n, id[top[x]], id[x])) % mod;
        x = f[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ret = (ret + query(1, 1, n, id[x], id[y])) % mod;
    return ret;
}
 
//------------------树链剖分结束-------------------//
 
 
int main()
{
    init();
    int m, r;
    scanf("%d%d%d%d", &n, &m, &r, &mod);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] %= mod;
    for (int i = 1; i < n; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v);
    }
    dfs1(r, 0, 1), dfs2(r, r);
    build(1, 1, n);
    while(m--)
    {
        int opt, x, y, z;
        scanf("%d", &opt);
        if(opt==1)
        {
            scanf("%d%d%d", &x, &y, &z);
            update2(x,y,z);
        }
        if(opt==2)
        {
            scanf("%d%d", &x, &y);
            int ans = query2(x,y);
            printf("%d
", ans);
        }
        if(opt==3)
        {
            scanf("%d%d", &x, &z);
            update(1, 1, n, id[x], id[x] + siz[x] - 1, z);
        }
        if(opt==4)
        {
            scanf("%d", &x);
            int ans = query(1, 1, n, id[x], id[x] + siz[x] - 1);
            printf("%d
", ans);
        }
    }
}

　　

然后就是一个简单的树链剖分的练习

树链剖分注意初始化和数组稍微要开大一点，不然容易re

https://vjudge.net/problem/CodeForces-343D

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <stack>
#include <map>
#include <string>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 5e5 + 10;
int f[maxn], dep[maxn], siz[maxn], son[maxn], rk[maxn], top[maxn], id[maxn];
int head[maxn], cnt, tot, sum[maxn * 4], lazy[maxn * 4], n;
struct node
{
	int v, nxt;
	node(int v=0,int nxt=0):v(v),nxt(nxt){}
}ex[maxn*4];
void init()
{
	memset(son, 0, sizeof(son));
	memset(head, -1, sizeof(head));
	cnt = tot = 0;
}

void push_up(int id)
{
	sum[id] = sum[id << 1] + sum[id << 1 | 1];
}

void build(int id,int l,int r)
{
	lazy[id] = -1,sum[id] = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(id << 1, l, mid);
	build(id << 1 | 1, mid + 1, r);
	push_up(id);
}

void push_down(int id,int len1,int len2)
{
	if (lazy[id] == -1) return;
	sum[id << 1] = len1 * lazy[id];
	sum[id << 1 | 1] = len2 * lazy[id];
	lazy[id << 1] = lazy[id << 1 | 1] = lazy[id];
	lazy[id] = -1;
}

void update(int id,int l,int r,int x,int y,int val)
{
	if(x<=l&&y>=r)
	{
		sum[id] = val * (r - l + 1);
		lazy[id] = val;
		return;
	}
	int mid = (l + r) >> 1;
	push_down(id, mid - l + 1, r - mid);
	if (x <= mid) update(id << 1, l, mid, x, y, val);
	if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val);
	push_up(id);
}

int query(int id,int l,int r,int pos)
{
	if (l == r) return sum[id];
	int mid = (l + r) >> 1;
	int ans = 0;
	push_down(id, mid - l + 1, r - mid);
	if (pos <= mid) ans += query(id << 1, l, mid, pos);
	else ans += query(id << 1 | 1, mid + 1, r, pos);
	return ans;
}

void add(int u,int v)
{
	ex[cnt] = node(v, head[u]);
	head[u] = cnt++;
	ex[cnt] = node(u, head[v]);
	head[v] = cnt++;
}


void dfs1(int u,int fa,int depth)
{
	f[u] = fa, dep[u] = depth, siz[u] = 1;
	for(int i=head[u];i!=-1;i=ex[i].nxt)
	{
		int v = ex[i].v;
		if (v == fa) continue;
		dfs1(v, u, depth + 1);
		siz[u] += siz[v];
		if (siz[v] > siz[son[u]]) son[u] = v;
	}
}

void dfs2(int u,int t)
{
	top[u] = t;//标记这个结点的顶端
	id[u] = ++tot;//标记这个结点的标号
	rk[tot] = u;//标号对应的结点
	if (!son[u]) return;
	dfs2(son[u], t);
	for(int i=head[u];i!=-1;i=ex[i].nxt)
	{
		int v = ex[i].v;
		if (v == u) continue;
		if (v != son[u] && v != f[u]) dfs2(v, v);
	}
}

void update1(int x,int y,int z)
{
	while(top[x]!=top[y])
	{
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		update(1, 1, n, id[top[x]], id[x], z);
		x = f[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	update(1, 1, n, id[x], id[y], z);
}


int main()
{
	init();
	int m;
	scanf("%d", &n);
	for(int i=1;i<n;i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		add(u, v);
	}
	dfs1(1, 0, 1), dfs2(1, 1);
	build(1, 1, n);
	scanf("%d", &m);
	while(m--)
	{
		int opt, v;
		scanf("%d%d", &opt, &v);
		if (opt == 1) update(1, 1, n, id[v], id[v] + siz[v] - 1, 1);
		if (opt == 2) update1(v, 1, 0);
		if (opt == 3) printf("%d
", query(1, 1, n, id[v]));
	}
	return 0;
}

　　

然后就是学校oj的一个也很裸的树链剖分

https://10.64.70.166/problem/1005

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <stack>
#include <map>
#include <string>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
int f[maxn], dep[maxn], siz[maxn], son[maxn], rk[maxn], top[maxn], ver[maxn];
int head[maxn], cnt, tot, n, a[maxn], m;
ll sum[maxn * 4], lazy[maxn * 4];
struct node
{
	int v, nxt;
	node(int v=0,int nxt=0):v(v),nxt(nxt){}
}ex[maxn*4];
void init()
{
	memset(son, 0, sizeof(son));
	memset(head, -1, sizeof(head));
	cnt = tot = 0;
}

void push_up(int id)
{
	sum[id] = sum[id << 1] + sum[id << 1 | 1];
}

void build(int id,int l,int r)
{
	lazy[id] = 0;
	if (l == r) {
		sum[id] = a[rk[l]];
		return;
	}
	int mid = (l + r) >> 1;
	build(id << 1, l, mid);
	build(id << 1 | 1, mid + 1, r);
	push_up(id);
}

void push_down(int id,int len1,int len2)
{
	if (lazy[id] == 0) return;
	sum[id << 1] += len1 * lazy[id];
	sum[id << 1 | 1] += len2 * lazy[id];
	lazy[id << 1] += lazy[id];
	lazy[id << 1 | 1] += lazy[id];
	lazy[id] = 0;
}

void update(int id,int l,int r,int x,int y,int val)
{
	if(x<=l&&y>=r)
	{
		sum[id] += val * (r - l + 1);
		lazy[id] += val;
		return;
	}
	int mid = (l + r) >> 1;
	push_down(id, mid - l + 1, r - mid);
	if (x <= mid) update(id << 1, l, mid, x, y, val);
	if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val);
	push_up(id);
}

ll query(int id,int l,int r,int pos)
{
	if (l == r) return sum[id];
	int mid = (l + r) >> 1;
	ll ans = 0;
	push_down(id, mid - l + 1, r - mid);
	if (pos <= mid) ans += query(id << 1, l, mid, pos);
	else ans += query(id << 1 | 1, mid + 1, r, pos);
	return ans;
}

void add(int u,int v)
{
	ex[cnt] = node(v, head[u]);
	head[u] = cnt++;
	ex[cnt] = node(u, head[v]);
	head[v] = cnt++;
}


void dfs1(int u,int fa,int depth)
{
	f[u] = fa, dep[u] = depth, siz[u] = 1;
	for(int i=head[u];i!=-1;i=ex[i].nxt)
	{
		int v = ex[i].v;
		if (v == fa) continue;
		dfs1(v, u, depth + 1);
		siz[u] += siz[v];
		if (siz[v] > siz[son[u]]) son[u] = v;
	}
}

void dfs2(int u,int t)
{
	top[u] = t;//标记这个结点的顶端
	ver[u] = ++tot;//标记这个结点的标号
	rk[tot] = u;//标号对应的结点
	if (!son[u]) return;
	dfs2(son[u], t);
	for(int i=head[u];i!=-1;i=ex[i].nxt)
	{
		int v = ex[i].v;
		if (v == u) continue;
		if (v != son[u] && v != f[u]) dfs2(v, v);
	}
}

void update1(int x,int y,int z)
{
	while(top[x]!=top[y])
	{
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		update(1, 1, n, ver[top[x]], ver[x], z);
		x = f[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	update(1, 1, n, ver[x], ver[y], z);
}

int opt[maxn], ux[maxn], vx[maxn], xx[maxn];
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
		for(int i=1;i<n;i++)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		dfs1(1, 0, 1), dfs2(1, 1);
		build(1, 1, n);
		for(int i=1;i<=m;i++)
		{
			scanf("%d", &opt[i]);
			if(opt[i]==1)
			{
				scanf("%d%d", &ux[i], &xx[i]);
				update(1, 1, n, ver[ux[i]], ver[ux[i]] + siz[ux[i]] - 1, xx[i]);
			}
			if(opt[i]==2)
			{
				scanf("%d%d%d", &ux[i], &vx[i], &xx[i]);
				update1(ux[i], vx[i], xx[i]);
			}
			if(opt[i]==3)
			{
				int t;
				scanf("%d", &t);
				if(opt[t]==1)
				{
					ux[i] = ux[t], xx[i] = -xx[t];
					update(1, 1, n, ver[ux[i]], ver[ux[i]] + siz[ux[i]] - 1, xx[i]);
				}
				if(opt[t]==2)
				{
					ux[i] = ux[t], vx[i] = vx[t], xx[i] = -xx[t];
					update1(ux[i], vx[i], xx[i]);
				}
			}
			if(opt[i]==4)
			{
				scanf("%d", &ux[i]);
				ll ans = query(1, 1, n, ver[ux[i]]);
				printf("%lld
", ans);
			}
			
		}
	}
}