【Vijos】lxhgww的奇思妙想（长链剖分）

题意：给定一棵n个点的树，m次强制在线的询问，每次询问x的k级祖先的编号

n<=3e5，m<=1.8e6

思路：参考资料：https://zhuanlan.zhihu.com/p/25984772

https://blog.bill.moe/long-chain-subdivision-notes/

不知道为什么写成f[v][0]=u会RE一部分，f[u][0]=fa就AC了，下次写长链剖分的时候注意

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef pair<int,int> PII;
  7 typedef pair<ll,ll> Pll;
  8 typedef vector<int> VI;
  9 typedef vector<PII> VII;
 10 typedef pair<ll,int>P;
 11 #define N  300010
 12 #define M  600010
 13 #define fi first
 14 #define se second
 15 #define MP make_pair
 16 #define pi acos(-1)
 17 #define mem(a,b) memset(a,b,sizeof(a))
 18 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 19 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 20 #define lowbit(x) x&(-x)
 21 #define Rand (rand()*(1<<16)+rand())
 22 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 23 #define ls p<<1
 24 #define rs p<<1|1
 25 
 26 const ll MOD=1e9+7,inv2=(MOD+1)/2;
 27       double eps=1e-6;
 28       ll INF=1ll<<62;
 29       ll inf=5e13;
 30       int dx[4]={-1,1,0,0};
 31       int dy[4]={0,0,-1,1};
 32 
 33 int f[N][20];
 34 vector<int> up[N],down[N];
 35 int head[M],vet[M],nxt[M],tot;
 36 int dep[M],son[M],top[M],len[M],hb[M],mx[M];
 37 
 38 int read()
 39 {
 40    int v=0,f=1;
 41    char c=getchar();
 42    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 43    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 44    return v*f;
 45 }
 46 
 47 void add(int a,int b)
 48 {
 49     nxt[++tot]=head[a];
 50     vet[tot]=b;
 51     head[a]=tot;
 52 }
 53 
 54 void dfs1(int u,int fa,int depth)
 55 {
 56     mx[u]=dep[u]=depth;
 57     f[u][0]=fa;
 58     rep(i,1,19) f[u][i]=f[f[u][i-1]][i-1];
 59     int e=head[u];
 60     while(e)
 61     {
 62         int v=vet[e];
 63         if(v!=fa)
 64         {
 65             dfs1(v,u,depth+1);
 66             if(mx[v]>mx[son[u]])
 67             {
 68                 son[u]=v;
 69                 mx[u]=mx[v];
 70             }
 71         }
 72         e=nxt[e];
 73     }
 74 }
 75 
 76 void dfs2(int u,int ance,int l)
 77 {
 78     len[u]=l;
 79     top[u]=ance;
 80     if(son[u])
 81     {
 82         dfs2(son[u],ance,l+1);
 83         len[u]=len[son[u]];
 84     }
 85     int e=head[u];
 86     while(e)
 87     {
 88         int v=vet[e];
 89         if(v!=f[u][0]&&v!=son[u]) dfs2(v,v,1);
 90         e=nxt[e];
 91     }
 92 }
 93 
 94 int query(int u,int k)
 95 {
 96     if(k>dep[u]) return 0;
 97     if(!k) return u;
 98     u=f[u][hb[k]];
 99     k-=(1<<hb[k]);
100     if(!k) return u;
101     if(dep[u]-dep[top[u]]==k) return top[u];
102     if(dep[u]-dep[top[u]]>k) return down[top[u]][dep[u]-dep[top[u]]-k-1];
103     return up[top[u]][k-(dep[u]-dep[top[u]])-1];
104 }
105 
106 int main()
107 {
108     int n=read();
109     rep(i,1,n) head[i]=0;
110     tot=0;
111     rep(i,1,n-1)
112     {
113         int x=read(),y=read();
114         add(x,y);
115         add(y,x);
116     }
117     dep[1]=0;
118     dfs1(1,0,1);
119     dfs2(1,1,1);
120     rep(i,1,19)
121      rep(j,(1<<(i-1)),((1<<i)-1)) hb[j]=i-1; //highbit
122     rep(i,1,n)
123      if(top[i]==i)
124      {
125          int k=i;
126          rep(j,1,len[i])
127          {
128              k=f[k][0];
129              up[i].push_back(k);
130          }
131          k=i;
132          rep(j,1,len[i])
133          {
134              k=son[k];
135              down[i].push_back(k);
136          }
137      }
138 
139     int m=read(),ans=0;
140     rep(i,1,m)
141     {
142         int x=read(),k=read();
143         x^=ans; k^=ans;
144         ans=query(x,k);
145         printf("%d\n",ans);
146     }
147     return 0;
148 }

 

posted on 2019-09-18 20:29  myx12345  阅读(169)  评论(0)  编辑  收藏  举报

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