Codeforces 384E 線段樹+dfs序

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題目連結：點擊開啟連結

題意：

給定n個點，m個詢問的無向樹（1為根）

下面n個數表示每個點的權值

下面n-1行給出樹

操作1：x點權值+v, x的第 i & 1 的兒子-v， 第 !(i&1) 的兒子+v

操作2：詢問x點權值

dfs把樹轉成序列

根據深度把點分成2組

分別用線段樹維護。。

然後Y一下

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<set>#include<queue>#include<vector>#include<map>using namespace std;#define ll __int64#define L(x) (x<<1)#define R(x) (x<<1|1)#define N 201000ll n ,m;inline ll Mid(ll a,ll b){return (a+b)>>1;}struct Edge{    ll from, to, nex;}edge[N<<1];ll head[N], edgenum;void add(ll u,ll v){    Edge E = {u,v,head[u]};    edge[edgenum] = E;    head[u] = edgenum++;}ll in[N], out[N], fa[N], Time, dep[N], V[N];void dfs(ll u, ll father, ll deep){    fa[u] = father;    dep[u] = deep;    in[u] = ++Time;    for(ll i = head[u];~i; i = edge[i].nex){        ll v = edge[i].to;      if(v==father)continue;        dfs(v,u,deep+1);    }    out[u] = Time;}struct node{    struct E{        ll l, r, val, lazy;    }t[N<<2];    void push_down(ll id){        if(t[id].l==t[id].r || t[id].lazy == 0)return ;        t[L(id)].val += t[id].lazy;        t[R(id)].val += t[id].lazy;        t[L(id)].lazy+=t[id].lazy;        t[R(id)].lazy+=t[id].lazy;        t[id].lazy = 0;    }    void build(ll l, ll r, ll id){        t[id].l = l; t[id].r = r;        t[id].val = 0;        t[id].lazy = 0;        if(l==r)return;        ll mid = Mid(l,r);        build(l,mid,L(id));build(mid+1,r,R(id));    }    void update(ll l, ll r,ll val,ll id){        push_down(id);        if(l == t[id].l && t[id].r == r) { t[id].val += val; t[id].lazy = val; return ;}                ll mid = Mid(t[id].l, t[id].r);        if(mid<l)update(l,r,val,R(id));        else if(r<=mid)update(l,r,val,L(id));        else {            update(l,mid,val,L(id));            update(mid+1,r,val,R(id));        }    }    ll query(ll l, ll r, ll id){        push_down(id);        if(l == t[id].l && t[id].r == r)return t[id].val;                ll mid = Mid(t[id].l, t[id].r);        if(mid<l)return query(l,r,R(id));        else if(r<=mid)return query(l,r,L(id));        return query(l,mid,L(id))+query(mid+1,r,R(id));    }}tree[2];ll query(ll u){    ll ans = tree[dep[u]&1].query(in[u],in[u],1);    return ans;}void init(){Time = 0; memset(head, -1, sizeof head); edgenum = 0;}int main(){      ll i, j, u, v;    while(cin>>n>>m) {        init();        for(i=1;i<=n;i++)cin>>V[i];        for(i=1;i<n;i++){            cin>>u>>v;            add(u,v);            add(v,u);        }        dfs(1,-1,1);        tree[0].build(1,n,1);           tree[1].build(1,n,1);        while(m--){            cin>>u;            if(u==1) {                cin>>u>>v;                tree[dep[u]&1].update(in[u],out[u],v,1);                if(in[u]<out[u])                    tree[!(dep[u]&1)].update(in[u]+1,out[u],-v,1);            }            else {                cin>>u;                cout<<query(u)+V[u]<<endl;            }        }    }    return 0;  }