BZOJ3589 动态树(树链剖分+容斥原理)

  显然容斥后转化为求树链的交。这个题非常良心的保证了查询的路径都是到祖先的,求交就很休闲了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
#define ui unsigned int
#define inf ((ui)4294967295)
#define p31 2147483647
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,p[N],fa[N],deep[N],son[N],size[N],top[N],dfn[N],L[N<<2],R[N<<2],u[6],v[6],flag[6],k,cnt,t;
ui tree[N<<2],lazy[N<<2],ans;
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs1(int k)
{
    size[k]=1;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k])
    {
        fa[edge[i].to]=k;
        deep[edge[i].to]=deep[k]+1;
        dfs1(edge[i].to);
        size[k]+=size[edge[i].to];
        if (size[edge[i].to]>size[son[k]]) son[k]=edge[i].to;
    }
}
void dfs2(int k,int from) 
{
    dfn[k]=++cnt;top[k]=from;
    if (son[k]) dfs2(son[k],from);
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]&&edge[i].to!=son[k]) dfs2(edge[i].to,edge[i].to);
}
void build(int k,int l,int r)
{
    L[k]=l,R[k]=r;
    if (l==r) return;
    int mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
}
void up(int k){tree[k]=tree[k<<1]+tree[k<<1|1];}
void update(int k,ui x){tree[k]+=(R[k]-L[k]+1)*x,lazy[k]+=x;}
void down(int k){update(k<<1,lazy[k]),update(k<<1|1,lazy[k]),lazy[k]=0;}
void add(int k,int l,int r,ui x)
{
    if (L[k]==l&&R[k]==r){update(k,x);return;}
    if (lazy[k]) down(k);
    int mid=L[k]+R[k]>>1;
    if (r<=mid) add(k<<1,l,r,x);
    else if (l>mid) add(k<<1|1,l,r,x);
    else add(k<<1,l,mid,x),add(k<<1|1,mid+1,r,x);
    up(k);
}
ui query(int k,int l,int r)
{
    if (L[k]==l&&R[k]==r) return tree[k];
    if (lazy[k]) down(k);
    int mid=L[k]+R[k]>>1;
    if (r<=mid) return query(k<<1,l,r);
    else if (l>mid) return query(k<<1|1,l,r);
    else return query(k<<1,l,mid)+query(k<<1|1,mid+1,r);
}
ui sum(int x,int y)
{
    ui ans=0;
    while (top[x]!=top[y])
    {
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        ans+=query(1,dfn[top[x]],dfn[x]);
        x=fa[top[x]];
    }
    if (deep[x]<deep[y]) swap(x,y);
    ans+=query(1,dfn[y],dfn[x]);
    return ans;
}
int lca(int x,int y)
{
    while (top[x]!=top[y])
    {
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        x=fa[top[x]];
    }
    if (deep[x]<deep[y]) swap(x,y);
    return y;
}
bool in(int x,int y){return dfn[x]<=dfn[y]&&dfn[x]+size[x]-1>=dfn[y];}
void calc(int op)
{
    int x=0,y=0;
    for (int i=1;i<=k;i++)
    if (flag[i])
    {
        if (!x) x=u[i],y=v[i];
        else
        {
            int p=u[i],q=v[i];
            if (deep[x]>deep[p]) swap(x,p),swap(y,q);
            if (in(x,p)&&in(p,y)) x=p,y=lca(y,q);
            else return;
        }
    }
    if (x==0) return;
    else if (op>0) ans+=sum(x,y);
    else ans+=inf-sum(x,y)+1;
}
void dfs(int cur,int op)
{
    if (cur>k) {calc(op);return;}
    flag[cur]=1;dfs(cur+1,-op);
    flag[cur]=0;dfs(cur+1,op);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3589.in","r",stdin);
    freopen("bzoj3589.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    dfs1(1);
    dfs2(1,1);
    build(1,1,n);
    int m=read();
    while (m--)
    {
        int op=read();
        if (op==0)
        {
            int x=read(),y=read();
            add(1,dfn[x],dfn[x]+size[x]-1,y);
        }
        if (op==1) 
        {
            k=read();ans=0;
            for (int i=1;i<=k;i++) u[i]=read(),v[i]=read();
            for (int i=1;i<=k;i++) if (dfn[u[i]]>dfn[v[i]]) swap(u[i],v[i]);
            dfs(1,-1);printf("%u\n",ans&p31);
        }
    }
    return 0;
}

 

 

posted @ 2019-01-18 22:14  Gloid  阅读(175)  评论(0编辑  收藏  举报