HDU 3966 树链剖分后线段树维护

题意:

　　一棵树,

　　操作1.$path(a,b)$之间的点权$+k$

　　操作2.单点查询

题解:

树链剖分即可,注意代码细节,双向映射

主要是记录一下板子

#include <string.h>
#include <stdio.h>
#include <algorithm>
#define endl '
'
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(ii,x) for(int ii=head[x];ii;ii=edge[ii].next)
using namespace std;
const int maxn=5e4+20,maxm=2e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
//head
int casn,n,m,k;
int num[maxn];
class segtree{public:
#define nd node[now]
#define ndl node[now<<1]
#define ndr node[now<<1|1]
  int node[maxn*4],n;
  int *mp;
	void maketree(int s,int t,int now){
			if(s==t){
				nd=num[mp[s]];
				return ;
			}else nd=0;
			maketree(s,(s+t)/2,now<<1);
			maketree((s+t)/2+1,t,now<<1|1);
	}
	void init(int nn,int *mps){
		n=nn;mp=mps;
		maketree(1,n,1);
	}
	void update(int s,int t,int x){
		update(s,t,x,1,n,1);
	}
	int query(int pos){
		return query(pos,1,n,1);
	}
	void update(int s,int t,int x,int l,int r,int now=1){
			if(s<=l&&t>=r) {
				nd+=x;
				return ;
			}
			ndl+=nd,ndr+=nd;nd=0;
			int mid=(l+r)/2;
			if(s<=mid) update(s,t,x,l,mid,now<<1);
			if(t>mid) update(s,t,x,mid+1,r,now<<1|1);
	}
	int query(int pos,int l,int r,int now=1){
			if(l==r) return nd;
			ndl+=nd,ndr+=nd;nd=0;
			int mid=(l+r)/2;
			if(pos<=mid) return query(pos,l,mid,now<<1);
			else  return query(pos,mid+1,r,now<<1|1);
	}
}tree;
namespace chain{
	struct data_e{
		int to,next;
	}edge[maxn<<1];
	int head[maxn],nume,mp[maxn];
	inline void addedge(int a,int b){
		edge[++nume]={b,head[a]};
		head[a]=nume;
	}
	int ltop[maxn],fa[maxn],deep[maxn];
	int sz[maxn],remp[maxn];
	int son[maxn],cnt;
	void init(){
    rep(i,1,n) head[i]=0;
    cnt=0,nume=1;
	}
  void dfs1(int now,int pre,int d){
		deep[now]=d,fa[now]=pre,sz[now]=1,son[now]=0;
		forn(i,now){
			int to=edge[i].to;
			if(to!=pre) {
				dfs1(to,now,d+1);
				sz[now]+=sz[to];
				if(sz[to]>sz[son[now]]) son[now]=to;
			}
		}
  }
  void dfs2(int now,int pre,int sp){
    ltop[now]=sp;mp[now]=++cnt;remp[cnt]=now;
		if(son[now])  dfs2(son[now],now,sp);
		forn(i,now){
			int to=edge[i].to;
			if(to!=son[now]&&to!=pre) dfs2(to,now,to);
		}
  }
  void update(int a,int b,int val){
    while(ltop[a]!=ltop[b]){
			if(deep[ltop[a]]<deep[ltop[b]])swap(a,b);
			tree.update(mp[ltop[a]],mp[a],val);
			a=fa[ltop[a]];
    }
		if(deep[a]>deep[b])swap(a,b);
		tree.update(mp[a],mp[b],val);
  }
};

int main() {
  while(~scanf("%d %d %d",&n,&m,&k)){
		rep(i,1,n) scanf("%d",num+i);
		chain::init();
    while(m--){
			int a,b;scanf("%d%d",&a,&b);
			chain::addedge(a,b);
			chain::addedge(b,a);
    }
    chain::dfs1(1,0,1);
    chain::dfs2(1,1,1);
		tree.init(n,chain::remp);
		while(k--){
			char x[10];int a,b,c;
			scanf("%s",x);
			if(x[0]=='Q'){
				scanf("%d",&a);
				printf("%d
",tree.query(chain::mp[a]));
			}else if(x[0]=='I'){
				scanf("%d %d %d",&a,&b,&c);
				chain::update(a,b,c);
			}else {
				scanf("%d %d %d",&a,&b,&c);
				chain::update(a,b,-c);
			}
		}
	}
}