前置知识
解法
树链剖分后维护一个支持区间修改,单点查询的线段树即可。
也可以树上差分,同 146. DFS 序 3,树上差分 1 的 \(1,2\) 操作,时间复杂度比树链剖分更优。
代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define sort stable_sort
#define endl '\n'
struct node
{
int nxt,to;
}e[100010];
int head[100010],c[100010],cc[100010],siz[100010],fa[100010],dep[100010],son[100010],top[100010],dfn[100010],cnt=0,tot=0;
struct SMT
{
struct SegmentTree
{
int l,r,sum,lazy;
}tree[200010];
int lson(int x)
{
return x*2;
}
int rson(int x)
{
return x*2+1;
}
void pushup(int rt)
{
tree[rt].sum=tree[lson(rt)].sum+tree[rson(rt)].sum;
}
void build(int rt,int l,int r)
{
tree[rt].l=l;
tree[rt].r=r;
tree[rt].lazy=0;
if(l==r)
{
tree[rt].sum=cc[l];
return;
}
int mid=(l+r)/2;
build(lson(rt),l,mid);
build(rson(rt),mid+1,r);
pushup(rt);
}
void pushdown(int rt)
{
if(tree[rt].lazy!=0)
{
tree[lson(rt)].sum+=tree[rt].lazy*(tree[lson(rt)].r-tree[lson(rt)].l+1);
tree[rson(rt)].sum+=tree[rt].lazy*(tree[rson(rt)].r-tree[rson(rt)].l+1);
tree[lson(rt)].lazy+=tree[rt].lazy;
tree[rson(rt)].lazy+=tree[rt].lazy;
tree[rt].lazy=0;
}
}
void update(int rt,int x,int y,int val)
{
if(x<=tree[rt].l&&tree[rt].r<=y)
{
tree[rt].sum+=val*(tree[rt].r-tree[rt].l+1);
tree[rt].lazy+=val;
return;
}
pushdown(rt);
int mid=(tree[rt].l+tree[rt].r)/2;
if(x<=mid)
{
update(lson(rt),x,y,val);
}
if(y>mid)
{
update(rson(rt),x,y,val);
}
}
int query(int rt,int x,int y)
{
if(x<=tree[rt].l&&tree[rt].r<=y)
{
return tree[rt].sum;
}
pushdown(rt);
int mid=(tree[rt].l+tree[rt].r)/2,ans=0;
if(x<=mid)
{
ans+=query(lson(rt),x,y);
}
if(y>mid)
{
ans+=query(rson(rt),x,y);
}
return ans;
}
}T;
void add(int u,int v)
{
cnt++;
e[cnt].nxt=head[u];
e[cnt].to=v;
head[u]=cnt;
}
void dfs1(int x,int father)
{
siz[x]=1;
fa[x]=father;
dep[x]=dep[father]+1;
for(int i=head[x];i!=0;i=e[i].nxt)
{
if(e[i].to!=father)
{
dfs1(e[i].to,x);
siz[x]+=siz[e[i].to];
son[x]=(siz[e[i].to]>siz[son[x]])?e[i].to:son[x];
}
}
}
void dfs2(int x,int father,int id)
{
top[x]=id;
tot++;
dfn[x]=tot;
cc[tot]=c[x];
if(son[x]!=0)
{
dfs2(son[x],x,id);
for(int i=head[x];i!=0;i=e[i].nxt)
{
if(e[i].to!=father&&e[i].to!=son[x])
{
dfs2(e[i].to,x,e[i].to);
}
}
}
}
void update1(int u,int v,int val)
{
while(top[u]!=top[v])
{
if(dep[top[u]]>dep[top[v]])
{
T.update(1,dfn[top[u]],dfn[u],val);
u=fa[top[u]];
}
else
{
T.update(1,dfn[top[v]],dfn[v],val);
v=fa[top[v]];
}
}
if(dep[u]<dep[v])
{
T.update(1,dfn[u],dfn[v],val);
}
else
{
T.update(1,dfn[v],dfn[u],val);
}
}
int query1(int pos)
{
return T.query(1,dfn[pos],dfn[pos]);
}
int main()
{
int t,n,m,u,v,w,i,j;
cin>>t;
for(j=1;j<=t;j++)
{
cnt=tot=0;
memset(e,0,sizeof(e));
memset(head,0,sizeof(head));
memset(son,0,sizeof(son));
cin>>n;
for(i=1;i<=n-1;i++)
{
cin>>u>>v;
u++;
v++;
add(u,v);
add(v,u);
}
dfs1(1,0);
dfs2(1,0,1);
T.build(1,1,n);
cin>>m;
for(i=1;i<=m;i++)
{
cin>>u>>v>>w;
u++;
v++;
update1(u,v,w);
}
cout<<"Case #"<<j<<":"<<endl;
for(i=1;i<=n;i++)
{
cout<<query1(i)<<endl;
}
}
return 0;
}
标签:rt,lazy,int,题解,Energy,tree,Lightning,100010,top
From: https://www.cnblogs.com/The-Shadow-Dragon/p/18213457