Gym - 101908L 树链剖分裸题

VJ的链接:https://cn.vjudge.net/problem/Gym-101908L 题目大意: 一棵$n$个点的树上，查询$a$ ~ $b$，$c$ ~ $d$两段路径公共的点的个数。

解题思路: 读完题就感觉是树链剖分的裸题呀... 把每个点的值初始化为$0$，对于每次询问，把$a$ ~ $b$这条路径的值$+1$，查询$c$ ~ $d$的区间和，再把$a$ ~ $b$这条路径的值$-1$改回来。$c$ ~ $d$的区间和就是交点数量。

代码: (就是树链剖分维护点权的板子，改了下main就过了...)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100000+10;
int w[maxn];
int N, Q;
struct
{
    int to,next;    
}e[maxn<<1];

int head[maxn],edgeNum;
void add(int u,int v)
{
    e[edgeNum].next = head[u];
    e[edgeNum].to = v;
    head[u] = edgeNum++;
}

/*-------------------------树剖------------------------------*/
int deep[maxn],fa[maxn],siz[maxn],son[maxn];
void dfs1(int u,int pre,int d)
{
    deep[u] = d;
    fa[u] = pre;
    siz[u] = 1;
    son[u] = 0;
    for(int i=head[u];~i;i=e[i].next)
    {
        int v = e[i].to;
        if(v!=pre)
        {
            dfs1(v,u,d+1);
            siz[u] += siz[v];
            if(siz[v]>siz[son[u]])
                son[u] = v;
        }
    }
}

int top[maxn],id[maxn],rk[maxn],cnt;
int &n = cnt;
void dfs2(int u,int t)
{
    top[u] = t;
    id[u] = ++cnt;
    rk[cnt] = u;
    if(!son[u]) return;

    dfs2(son[u],t);

    for(int i=head[u];~i;i=e[i].next)
    {
        int v = e[i].to;
        if(v!=son[u]&&v!=fa[u])
            dfs2(v,v);
    }
}
/*-------------------------树剖------------------------------*/

/*-------------------------线段树------------------------------*/
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[maxn<<2],lazy[maxn<<2];
void pushup(int rt)
{
    sum[rt] = (sum[rt<<1] + sum[rt<<1|1]);
}

void build(int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt] = w[rk[l]];
        return ;
    }
    int m = l+r>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void pushdown(int rt,int l,int r)
{
    if(lazy[rt])
    {
        lazy[rt<<1] = (lazy[rt<<1] + lazy[rt]);
        lazy[rt<<1|1] = (lazy[rt<<1|1] + lazy[rt]);
        sum[rt<<1] += (lazy[rt] * l);
        sum[rt<<1|1] += (lazy[rt] * r);
        lazy[rt] = 0;
    }
}

void update(int L,int R,int val,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        sum[rt] = (sum[rt] + (val) * ((r-l+1)));
        lazy[rt] += val;
        return ;
    }
    int m = l + r >> 1;
    pushdown(rt,m-l+1,r-m);
    if(L<=m)
        update(L,R,val,lson);
    if(R>m)
        update(L,R,val,rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
        return sum[rt];
    int m = l + r >> 1,ans = 0;
    pushdown(rt,m-l+1,r-m);
    if(L<=m)
        ans = (ans + query(L,R,lson));
    if(R>m)
        ans = (ans + query(L,R,rson));
    return ans;
}
/*-------------------------线段树------------------------------*/

/*-----------------------树剖加线段树--------------------------*/
void update(int x,int y,int z)
{
    while(top[x]!=top[y])
    {
        if(deep[top[x]]<deep[top[y]])
            swap(x,y);
        update(id[top[x]],id[x],z,1,n,1);
        x = fa[top[x]];
    }
    if(deep[x]>deep[y])
        swap(x,y);
    update(id[x],id[y],z,1,n,1);
}

int query(int x,int y)
{
    int ans = 0;
    while(top[x] != top[y])
    {
        if(deep[top[x]] < deep[top[y]])
            swap(x,y);
        ans = (ans + query(id[top[x]],id[x],1,n,1));
        x = fa[top[x]];
    }
    if(deep[x]>deep[y])
        swap(x,y);
    ans = (ans + query(id[x],id[y],1,n,1));
    return ans;
}
/*-----------------------树剖加线段树--------------------------*/

void init()
{
    memset(head,-1,4*N+4);
    cnt = edgeNum = 0;
}

int u, v, x1, y1, x2, y2;
int main()
{   
   	scanf("%d%d",&N,&Q);
   	init();
   	for(int i=1;i<N;++i)
	{
		w[i] = 0;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs1(1,0,0);
	dfs2(1,1);
	build(1,n,1);
	while(Q--)
	{
		scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
		update(x1,y1,1);
		printf("%d\n",query(x2,y2));
		update(x1,y1,-1);
	}
    return 0;
}