省选D2T3考板子可真是不多见呢。。。。~~~
这题就是一个裸的树链剖分,对于每一个Add操作,维护从u至v的路径,对于每一个Query操作,询问以u为根的子树之和。如果不会树链剖分可以看我的往期博客,具体细节在代码之中就不多赘述了~
下面给出参考代码:
#include<iostream>
#include<cstdio>
#define N 400005
#define M 800005
#define lc k*2
#define rc k*2+1
#define mid (l+r)/2
#define int long long
using namespace std;
struct node
{
int l,r,w,tag;
}tree[4*N];
int n,m,r,p,x,y,z;
int v[M],head[M],nxt[M],cnt;
int dep[N],fa[N],son[N],size[N],top[N],seg[N],id;
char q;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-48;ch=getchar();}
return x*f;
}
void add(int a,int b)
{
v[++cnt]=b;
nxt[cnt]=head[a];
head[a]=cnt;
}
void dfs1(int node,int father)
{
dep[node]=dep[father]+1;
fa[node]=father;
size[node]=1;
int maxson=-1;
for(int i=head[node];i;i=nxt[i])
{
int go=v[i];
if(go==father)continue;
dfs1(go,node);
size[node]+=size[go];
if(size[go]>maxson)maxson=size[go],son[node]=go;
}
}
void dfs2(int node,int topfather)
{
seg[node]=++id;
top[node]=topfather;
if(!son[node])return;
dfs2(son[node],topfather);
for(int i=head[node];i;i=nxt[i])
{
int go=v[i];
if(go==fa[node]||go==son[node])continue;
dfs2(go,go);
}
}
void build(int l,int r,int k)
{
tree[k].l=l;tree[k].r=r;
if(l==r)return;
build(l,mid,lc);
build(mid+1,r,rc);
}
void pushdown(int k)
{
tree[lc].tag+=tree[k].tag;
tree[rc].tag+=tree[k].tag;
tree[lc].w+=(tree[lc].r-tree[lc].l+1)*tree[k].tag;
tree[rc].w+=(tree[rc].r-tree[rc].l+1)*tree[k].tag;
tree[k].tag=0;
return;
}
void pushup(int k)
{
tree[k].w=tree[lc].w+tree[rc].w;
return;
}
void add(int x,int y,int k)
{
int l=tree[k].l,r=tree[k].r;
if(l>=x&&r<=y)
{
tree[k].tag+=z;
tree[k].w+=(r-l+1)*z;
return;
}
if(tree[k].tag)pushdown(k);
if(x<=mid)add(x,y,lc);
if(y>mid)add(x,y,rc);
pushup(k);
}
int query(int x,int y,int k)
{
int l=tree[k].l,r=tree[k].r;
if(l>=x&&r<=y)
{
return tree[k].w;
}
if(tree[k].tag)pushdown(k);
int res=0;
if(x<=mid)res+=query(x,y,lc);
if(y>mid)res+=query(x,y,rc);
return res;
}
void Tadd(int x,int y)
{
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
add(seg[top[x]],seg[x],1);
x=fa[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
add(seg[x],seg[y],1);
}
int Treequery(int x)
{
return query(seg[x],seg[x]+size[x]-1,1);
}
signed main()
{
n=read();
for(int i=1;i<n;i++)
{
x=read();y=read();
x++;y++;
add(x,y);add(y,x);
}
dfs1(1,0);dfs2(1,1);
build(1,n,1);
m=read();
while(m--)
{
cin>>q;
if(q==‘A‘)
{
x=read();y=read();z=read();
x++;y++;
Tadd(x,y);
}
else
{
x=read();
x++;
cout<<Treequery(x)<<endl;
}
}
}
原文:https://www.cnblogs.com/szmssf/p/11145035.html