zoukankan      html  css  js  c++  java
  • bzoj 2588: Spoj 10628. Count on a tree

    DFS序在树上建出主席树,然后。。。。。

      1 #include<cstdio>
      2 #include<iostream>
      3 #include<cstring>
      4 #include<cstdlib>
      5 #include<cmath>
      6 #include<algorithm>
      7 #define ll long long
      8 #define M 200009
      9 using namespace std;
     10 ll read()
     11 {
     12   ll x=0,f=1;
     13   char ch=getchar();
     14   for(;ch<'0'||ch>'9';ch=getchar())
     15     if(ch=='-')
     16       f=-1;
     17   for(;ch>='0'&&ch<='9';ch=getchar())
     18     x=x*10+ch-'0';
     19   return x*f;
     20 }
     21 int lc[M][20],head[M],next[M],u[M],cnt,n,m,v[M],v1[M],n1,deep[M],pos[M],num[M],T,root[M];
     22 int tot,lastans;
     23 struct data
     24 {
     25     int l,r,sum;
     26 }a[20*M];
     27 void jia(int a1,int a2)
     28 {
     29     next[++cnt]=head[a1];
     30     head[a1]=cnt;
     31     u[cnt]=a2;
     32 }
     33 void dfs(int x)
     34 {
     35     num[++T]=x;
     36     pos[x]=T;
     37     for(int i=1;(1<<i)<=deep[x];i++)
     38         lc[x][i]=lc[lc[x][i-1]][i-1];
     39     for(int i=head[x];i;i=next[i])
     40       if(u[i]!=lc[x][0])
     41       {
     42           deep[u[i]]=deep[x]+1;
     43           lc[u[i]][0]=x;
     44           dfs(u[i]);
     45       }
     46 }
     47 void geng(int l,int r,int x,int &y,int z)
     48 {
     49     y=++tot;
     50     a[y].sum=a[x].sum+1;
     51     if(l==r)
     52       return;
     53     a[y].l=a[x].l;
     54     a[y].r=a[x].r;
     55     int mid=(l+r)>>1;
     56     if(mid>=z)
     57         geng(l,mid,a[x].l,a[y].l,z);
     58     else
     59         geng(mid+1,r,a[x].r,a[y].r,z);
     60 }
     61 int lca(int a1,int a2)
     62 {
     63     if(deep[a1]<deep[a2])
     64         swap(a1,a2);
     65     int a3=deep[a1]-deep[a2];
     66     for(int i=0;i<=16;i++)
     67         if(a3&(1<<i))
     68           a1=lc[a1][i];
     69     for(int i=16;i>=0;i--)
     70         if(lc[a1][i]!=lc[a2][i])
     71       {
     72           a1=lc[a1][i];
     73           a2=lc[a2][i];
     74       }
     75     if(a1==a2)
     76         return a1;
     77     return lc[a1][0];
     78 }
     79 int cha(int u,int v,int t,int k)
     80 {
     81     int a1=root[pos[u]],a2=root[pos[v]],a3=root[pos[t]],a4=root[pos[lc[t][0]]];
     82     int l=1,r=n1-1,ans;
     83     for(;;)
     84       {
     85           if(l==r)
     86               return v1[l];
     87           int mid=(l+r)>>1;
     88           int a5=a[a[a1].l].sum+a[a[a2].l].sum-a[a[a3].l].sum-a[a[a4].l].sum;
     89           if(a5>=k)
     90             {
     91                 r=mid;
     92                 a1=a[a1].l;
     93                 a2=a[a2].l;
     94                 a3=a[a3].l;
     95                 a4=a[a4].l;
     96             }
     97           else
     98             {
     99                 k-=a5;
    100                 l=mid+1;
    101                 a1=a[a1].r;
    102                 a2=a[a2].r;
    103                 a3=a[a3].r;
    104                 a4=a[a4].r;
    105             }
    106       }
    107 }
    108 int main()
    109 {
    110     n=read();
    111     m=read(); 
    112     for(int i=1;i<=n;i++)
    113       v1[i]=v[i]=read();
    114     sort(v1+1,v1+n+1);
    115     n1=unique(v1+1,v1+n+1)-v1;
    116     for(int i=1;i<=n;i++)
    117       v[i]=lower_bound(v1+1,v1+n1,v[i])-v1;
    118     for(int i=1;i<n;i++)
    119       {
    120           int a1=read(),a2=read();
    121           jia(a1,a2);
    122           jia(a2,a1);
    123       }
    124     dfs(1);
    125     for(int i=1;i<=n;i++)
    126       geng(1,n1-1,root[pos[lc[num[i]][0]]],root[i],v[num[i]]);
    127     for(int i=1;i<=m;i++)
    128       {
    129           int u=read()^lastans,v=read(),k=read();
    130           int t=lca(u,v);
    131           printf("%d",lastans=cha(u,v,t,k));
    132           if(i!=m)
    133             printf("
    ");
    134       }
    135     return 0;
    136 }
  • 相关阅读:
    进制转换
    01背包基础
    6-14 Inspector s Dilemma uva12118(欧拉道路)
    9-4 Unidirectional TSP uva116 (DP)
    8-4 奖品的价值 uva11491(贪心)
    9-1 A Spy in the Metro uva1025 城市里的间谍 (DP)
    8-3 Bits Equalizer uva12545
    8-2 Party Games uva1610 (贪心)
    自动发邮件功能
    窗口截图.py
  • 原文地址:https://www.cnblogs.com/xiw5/p/5618481.html
Copyright © 2011-2022 走看看