给定一棵以1为根的树,支持两种操作:给某个点打上标记,询问某个点最近的有标记的祖先
比板子还裸,每次打标记就意味着把这个节点以及它的所有子树的值设为当前节点的编号
甚至不需要up操作,只需要把当前标记下传即可,下传时候注意比较标记与当前值的编号
由于是找最近的祖先,应当确保值的深度最大
此外build操作时候把所有节点的值设定为1即可,因为默认1是有标记的
代码:
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read() { char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x) { if(x<0)putchar(‘-‘),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int head[maxn],to[maxn],e[maxn];//Graph int sz[maxn],fa[maxn],dep[maxn],son[maxn];//using for d1 int ord[maxn],last[maxn];//using for d2 int tg,td,tt;//three timers,using for label graph,dfs and segment tree struct segment_tree{ int l,r,v,tag,ls,rs; }tr[maxn];//segment tree void add(int x,int y){ e[++tg]=y; to[tg]=head[x]; head[x]=tg; } void d1(int x){ sz[x]=1; for(int i=head[x];i;i=to[i]){ int y=e[i]; if(y==fa[x]) continue; fa[y]=x; dep[y]=dep[x]+1; d1(y); sz[x]+=sz[y]; if(sz[y]>sz[son[x]]) son[x]=y; } } void d2(int x){ ord[x]=++td; if(son[x]) d2(son[x]); for(int i=head[x];i;i=to[i]){ int y=e[i]; if(y!=fa[x]&&y!=son[x]) d2(y); } last[x]=td; } void down(int p){ if(tr[p].tag!=0){ if(dep[tr[p].tag]>dep[tr[tr[p].ls].v]){ tr[tr[p].ls].v=tr[p].tag; tr[tr[p].ls].tag=tr[p].tag; } if(dep[tr[p].tag]>dep[tr[tr[p].rs].v]){ tr[tr[p].rs].v=tr[p].tag; tr[tr[p].rs].tag=tr[p].tag; } tr[p].tag=0; } } void build(int x,int y){ ++tt; tr[tt].l=x; tr[tt].r=y; tr[tt].v=1; tr[tt].tag=0; if(x==y){ tr[tt].ls=tr[tt].rs=0; return; } else{ int tmp=tt,m=mm(x,y); tr[tmp].ls=tt+1,build(x,m); tr[tmp].rs=tt+1,build(m+1,y); } } void modify(int p,int x,int y,int z){ if(x<=tr[p].l&&tr[p].r<=y){ if(dep[z]>dep[tr[p].v]){ tr[p].v=z; tr[p].tag=z; } return; } down(p); int m=mm(tr[p].l,tr[p].r); if(x<=m) modify(tr[p].ls,x,y,z); if(y>m) modify(tr[p].rs,x,y,z); } int ask(int p,int pos){ if(tr[p].l==tr[p].r) return tr[p].v; int m=mm(tr[p].l,tr[p].r); down(p); if(pos<=m) return ask(tr[p].ls,pos); else return ask(tr[p].rs,pos); } void modify1(int p,int x,int y){ modify(1,ord[x],last[x],y); } int ask1(int p,int x){ return ask(1,ord[x]); } int n,m,x,y,q; char c; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>m; re(i,1,n-1){ cin>>x>>y; add(x,y); add(y,x); } d1(1); d2(1); build(1,n); re(i,1,m){ cin>>c>>q; if(c==‘C‘) modify1(1,q,q); else if(c==‘Q‘) cout<<ask1(1,q)<<endl; } return 0; }
P4092 [HEOI2016/TJOI2016]树(树链剖分)
原文:https://www.cnblogs.com/oneman233/p/11488998.html