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  • P2590 [ZJOI2008]树的统计

    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;
    }
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  • 原文地址:https://www.cnblogs.com/acm1ruoji/p/11904992.html
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