"Poj2104-for the K-large number of intervals (not modified)" Chairman Tree/Sustainable segment tree

The first chairman of the tree ~ however is the Tao compared to the water ... Because it doesn't have to be modified ...

Reproduced a Let me understand the chairman of the tree to explain it: http://blog.csdn.net/regina8023/article/details/41910615 (unauthorized, infringement delete)

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So what if you ask the number of times between i-j?

Because each segment of the tree line is the same, it is required that the number of occurrences of the numbers between l-r as long as the number of occurrences of the first R bit minus the number of occurrences of the first l-1 , is ans

But what if there is a modification operation?

If you follow the above practice, then the modification operation is O (NLOGN), the query is O (1), the modification will take a long time ...

Prefix and Lenovo to the tree-like array , then the prefix and the tree-like array to maintain the modified is O (LOGN*LOGN), Query O (logn), the query time is longer, but the change of time to shorten a lot!!

Attention:

Array of functional segment trees to open up a bit!!

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This problem is the template problem, first discrete, to find the interval K big time lx=root[l-1],rx=root[r], both sides at the same time to continue to make poor, look at the number of left children, if K is more large to reduce the left child to have the right child to find.

Code:

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <cmath>5#include <iostream>6#include <algorithm>7 using namespacestd;8 9 Const intn=100010, inf= (int) 1e9+ -;Ten structtrnode{ One     intlc,rc,cnt; A}t[ -*N]; - structnode{ -     intD,id; the }p[n]; - intn,m,tl,mx; - intA[n],num[n],root[n]; -  + BOOLCMP (node X,node y) {returnx.d<Y.d;} -  + intBtintLintR) A { at     intx=++tl; -     //a[x].l=l;a[x].r=r; -T[x].lc=t[x].rc=0; -T[x].cnt=0; -     if(l<R) -     { in         intMid= (L+R)/2; -T[x].lc=bt (L,MID); toT[X].RC=BT (mid+1, R); +     } -     returnx; the } *  $ intAddintRtintx)Panax Notoginseng { -     intnow=++tl,tmp=Now ; thet[now].cnt=t[rt].cnt+1; +     intL=1, r=Mx,mid; A      while(l<R) the     { +Mid= (L+R)/2; -         if(x<=mid) $         { $t[now].lc=++tl; -T[now].rc=t[rt].rc; -rt=t[rt].lc; thenow=tl; -R=mid;Wuyi         } the         Else -         { WuT[now].lc=t[rt].lc; -t[now].rc=++tl; Aboutrt=t[rt].rc; $now=tl; -L=mid+1; -         } -t[now].cnt=t[rt].cnt+1; A     } +     returntmp; the } -  $ intQueryintLxintRxintk) the { the     intL=1, r=Mx,mid; the      while(l<R) the     { -Mid= (L+R)/2; in         if(t[t[rx].lc].cnt-t[t[lx].lc].cnt>=k) the         { theR=mid; Aboutlx=t[lx].lc; therx=t[rx].lc; the         } the         Else +         { -L=mid+1; thek-=t[t[rx].lc].cnt-t[t[lx].lc].cnt;Bayilx=t[lx].rc; therx=t[rx].rc; the         } -     } -     returnl; the } the  the voidOutputintx) the { -printf"x =%d LC =%d rc =%d cnt =%d\n", x,t[x].lc,t[x].rc,t[x].cnt); the     if(T[X].LC) output (T[X].LC); the     if(t[x].rc) output (t[x].rc); the }94  the intMain () the { theFreopen ("a.in","R", stdin);98      while(SCANF ("%d%d", &n,&m)! =EOF) About     { -Tl=0; mx=0;101          for(intI=1; i<=n;i++)102         {103scanf"%d",&a[i]);104P[i].d=a[i];p [i].id=i; the         }106Sort (p+1, p+1+n,cmp);107p[0].d=INF;108          for(intI=1; i<=n;i++)109         { the             if(p[i].d!=p[i-1].D) mx++,num[mx]=p[i].d;111a[p[i].id]=MX; the         }113root[0]=BT (1, MX); the          for(intI=1; i<=n;i++) theRoot[i]=add (root[i-1],a[i]); the         //output (root[5]);117          for(intI=1; i<=m;i++)118         {119             intl,r,k; -scanf"%d%d%d",&l,&r,&k);121printf"%d\n", Num[query (root[l-1],root[r],k)]);122         }123     }124     return 0; the}

"Poj2104-for the K-large number of intervals (not modified)" Chairman Tree/Sustainable segment tree