P2590 [ZJOI2008]树的统计

Posted 2020-11-22 acm1ruoji

题目链接：

树链剖分裸题，可以当作熟悉模板写一下把

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;
#define ll long long
#define re register
#define pb push_back
#define fi first
#define se second
const int N=1e6+10;
const int mod=998244353;
void read(int &a)
{
    a=0;int d=1;char ch;
    while(ch=getchar(),ch>‘9‘||ch<‘0‘)
        if(ch==‘-‘)
            d=-1;
    a=ch^48;
    while(ch=getchar(),ch>=‘0‘&&ch<=‘9‘)
        a=(a<<3)+(a<<1)+(ch^48);
    a*=d;
}
struct note{int l,r;ll lazy,sum,ma;}tree[N<<1];
int top[N],id[N],rk[N],son[N],siz[N],f[N],cnt,dep[N],val[N],rt=1;
vector <int> v[N];
void dfs1(int x)
{
    siz[x]=1,dep[x]=dep[f[x]]+1;
    for(auto i:v[x])
    {
        if(i!=f[x])
        {
            f[i]=x,dfs1(i);
            siz[x]+=siz[i];
            if(siz[son[x]]<siz[i]) son[x]=i;
        }
    }
}
void dfs2(int x,int tp)
{
    top[x]=tp;id[x]=++cnt;rk[cnt]=x;
    if(son[x]) dfs2(son[x],tp);
    for(auto i:v[x]) if(i!=son[x]&&i!=f[x]) dfs2(i,i);
}
void build(int l,int r,int now)
{
    tree[now].l=l,tree[now].r=r;
    if(l==r) {tree[now].sum=val[rk[l]],tree[now].lazy=0,tree[now].ma=val[rk[l]];return;}
    int m=l+r>>1;
    build(l,m,now<<1),build(m+1,r,now<<1|1);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
    tree[now].ma=max(tree[now<<1].ma,tree[now<<1|1].ma);
}
void work(int now,int k)
{
    tree[now].sum=1ll*(tree[now].r-tree[now].l+1)*k;
    tree[now].ma=k;
    tree[now].lazy=k;
}
void pushdown(int now)
{
    work(now<<1,tree[now].lazy);
    work(now<<1|1,tree[now].lazy);
    tree[now].lazy=0;
}
void modify(int l,int r,int now,int k)
{
    if(tree[now].l>=l&&tree[now].r<=r) {work(now,k);return;}
    if(tree[now].lazy) pushdown(now);
    int m=tree[now].l+tree[now].r>>1;
    if(m>=l) modify(l,r,now<<1,k);
    if(m<r) modify(l,r,now<<1|1,k);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
    tree[now].ma=max(tree[now<<1].ma,tree[now<<1|1].ma);
}
ll query(int l,int r,int now)
{
    ll ans=0;
    if(tree[now].l>=l&&tree[now].r<=r) {return tree[now].sum;}
    if(tree[now].lazy) pushdown(now);
    int m=tree[now].l+tree[now].r>>1;
    if(m>=l) ans+=query(l,r,now<<1);
    if(m<r) ans+=query(l,r,now<<1|1);
    return ans;
}
ll queryy(int l,int r,int now)
{
    ll ans=-1e18;
    if(tree[now].l>=l&&tree[now].r<=r) {return tree[now].ma;}
    if(tree[now].lazy) pushdown(now);
    int m=tree[now].l+tree[now].r>>1;
    if(m>=l) ans=max(ans,queryy(l,r,now<<1));
    if(m<r) ans=max(ans,queryy(l,r,now<<1|1));
    return ans;
}
ll query1(int x,int y)
{
    ll ans=0;
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        ans+=query(id[top[x]],id[x],1);
        x=f[top[x]];
    }
    if(dep[x]>dep[y]) swap(x,y);
    return ans+query(id[x],id[y],1);
}
ll query2(int x,int y)
{
    ll ans=-1e18;
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        ans=max(ans,queryy(id[top[x]],id[x],1));
        x=f[top[x]];
    }
    if(dep[x]>dep[y]) swap(x,y);
    return ans=max(ans,queryy(id[x],id[y],1));
}
int main()
{
    int n,m;
    read(n);
    for(re int i=1,x,y;i<n;i++) read(x),read(y),v[x].pb(y),v[y].pb(x);
    for(re int i=1;i<=n;i++) read(val[i]);
    dfs1(rt),dfs2(rt,rt),build(1,n,1);
    read(m);
    char op[10];
    for(re int i=1,l,r;i<=m;i++)
    {
        scanf("%s %d %d",&op,&l,&r);
        if(op[1]==‘M‘) printf("%lld
",query2(l,r));
        else if(op[1]==‘S‘) printf("%lld
",query1(l,r));
        else modify(id[l],id[l],1,r);
    }
    return 0;
}