"Hihocoder 1167" Advanced Theoretical Computer Science (intersection of tree chains, segment tree or tree array maintenance interval and)

Test instructions

time limit: 20000ms single point time limit: 1000ms memory limit: 256MB description

Girly Fragrance These days are learning advanced theoretical computer science, but she has not learned anything, very painful. So she went out and flash and made some meaningless things to relax herself.
There is a tree of n nodes in front of the door, the fragrance found that there are N elves on the tree. However, these elves are shy and will only be active on a specific path. The Pokemon will be active on the AI to bi path.
Two elves are friends, when and only if their paths are of a common point.
So the fragrance wants to know, how many pairs of elves A and b,a and B are friends? Where a is not equal to b,a,b and b,a as a pair.

Input

The first row N and P (1 <= N, p <=100000) represent the size of the tree and the number of sprites. The nodes of the tree are labeled from 1 to N.
Next n-1 line, two numbers per line, a, a, indicates an edge between A and B.
Next P line, line I two number Ai,bi, indicating that the activity range of Elf I is AI to bi, where AI is not equal to BI.

Output

A row of answers that represents the logarithm.

Sample input

6 31 22 32 44 54 61 31 55 6

Sample output

2

Analysis

ORZ ...

I'm so stupid. Always want to cut the tree and the intersection of the line ... "And not a line segment

Great God's main puzzle here:

Two tree chains intersect, when and only if the LCA of a tree chain is on another tree chain, for each tree chain, count how many of the tree chains of LCA are included in him, sometimes two tree chains satisfy this condition, but only when the LCA of the two tree chains is equal, so it's a special sentence.

The above is clear, the interval and maintenance with a tree-like array can be.

1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <iostream>5#include <algorithm>6#include <cmath>7 using namespacestd;8 #defineMAXN 1000109 #defineLL Long LongTen  One structnode A { -     intX,y,next; -}t[maxn*2];intLen; the  - intFIRST[MAXN],PX[MAXN],PY[MAXN]; -  - voidInsintXinty) + { -t[++len].x=x;t[len].y=y; +t[len].next=first[x];first[x]=Len; A } at  - intSON[MAXN],DFN[MAXN],SM[MAXN],DEP[MAXN],FA[MAXN]; - voidDFS1 (intXintf) - { -sm[x]=1; son[x]=0;d ep[x]=dep[f]+1; fa[x]=F; -      for(intI=first[x];i;i=t[i].next)if(t[i].y!=f) in     { -         inty=t[i].y; to DFS1 (y,x); +sm[x]+=Sm[y]; -         if(Sm[y]>sm[son[x]]) son[x]=y; the     } * } $ Panax Notoginseng inttp[maxn],cnt; - voidDFS2 (intXintFintTPP) the { +dfn[x]=++cnt;tp[x]=TPP; A     if(Son[x]) DFS2 (SON[X],X,TPP); the      for(intI=first[x];i;i=t[i].next)if(t[i].y!=f&&t[i].y!=Son[x]) + DFS2 (T[I].Y,X,T[I].Y); - } $  $ intC[maxn],n; - BOOLLCA[MAXN]; -  the voidAddintXinty) - {Wuyi      for(inti=x;i<=n;i+=i& (-i)) thec[i]+=y; - } Wu  - intQueryintLintR) About { $     intans=0; -      for(inti=r;i>=1;i-=i& (-i)) -ans+=C[i]; -l--; A      for(inti=l;i>=1;i-=i& (-i)) +ans-=C[i]; the     returnans; - } $  the intGans (intXintYintp) the { the     intans=0, TT; the      while(tp[x]!=Tp[y]) -     { in         if(Dep[tp[x]]<dep[tp[y]]) tt=x,x=y,y=tt; the         if(p==1) ans+=query (dfn[tp[x]],dfn[x]); thex=Fa[tp[x]]; About     } the     if(Dep[x]<dep[y]) tt=x,x=y,y=tt; the     if(p==1) the     { +ans+=query (dfn[y],dfn[x]); -         returnans; the     }Bayi     Else returny; the } the  - intMain () - { the     intp; theLL ans=0; thescanf"%d%d",&n,&p); thelen=0; -memset (First,0,sizeof(first)); the      for(intI=1; i<n;i++) the     { the         intx, y;94scanf"%d%d",&x,&y); the ins (x, y); ins (y,x); the     } the      for(intI=1; i<=p;i++) scanf ("%d%d",&px[i],&py[i]);98sm[0]=0;d ep[0]=0; AboutDFS1 (1,0); cnt=0; -DFS2 (1,0,1);101Memset (c,0,sizeof(c));102memset (LCA,0,sizeof(LCA));103      for(intI=1; i<=p;i++)104     { the         intX=gans (Px[i],py[i],0);106lca[x]=1;107Add (Dfn[x],1);108     }109      for(intI=1; i<=p;i++) the     {111         intX=gans (Px[i],py[i],1); theans+=x;113     } the      for(intI=1; i<=n;i++)if(Lca[i]) the     { the         intx=query (Dfn[i],dfn[i]);117ans-=x* (x1)/2+x;118     }119printf"%lld\n", ans); -     return 0;121}

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2016-11-10 18:17:12

"Hihocoder 1167" Advanced Theoretical Computer Science (intersection of tree chains, segment tree or tree array maintenance interval and)