A study record of point division in a tree

Source: Internet
Author: User

Follow the engine room of the trend of the point of Division, found that this thing is actually very good to write, learn to think, it is easy to yy out. directly on the exercise.
Point division of the template problem, the first place x To the current subtree and r o o t Span style= "Display:inline-block; width:0px; Height:2.563em; " > The distance is g x , it satisfies g x + g y ≤k You can add the answer, but note that if x y Span style= "Display:inline-block; width:0px; Height:2.723em; " > In the same subtrees tree, it is necessary to delete the contribution to the answer, thinking x y Span style= "Display:inline-block; width:0px; Height:2.723em; " > is calculated once at the China in which it resides. Note that there is no root tree to the root tree process, you need to select the tree's center of gravity to prevent complexity from O(N log 2 N) Back to o ( n 2 ) Code

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define INF 100000Using namespace Std;struct hp{intU,v,w;} a[20001];struct hq{int size;}tree[10001];bool done[10001];intpoint[10001],Next[20001];intlen[10001];intNm, e=0, t=0;intsize,root,sizenow,tot,ans=0; void Add (intUintVintW) {e++; a[e].u=u; a[e].v=v; a[e].w=w;Next[E]=point[u];    Point[u]=e; e++; A[e].u=v; A[e].v=u; A[e].w=w;Next[E]=point[v]; Point[v]=e;} void Findsize (intNowint  Last){inti,tmp=0; sizenow++; for(i=point[now];i;i=Next[i])if(a[i].v!= Last&AMP;&AMP;!DONE[A[I].V]) findsize (A[i].v,now);} void Findroot (intNowint  Last){inti,tmp=0; Tree[now].size=1; tmp=0; for(i=point[now];i;i=Next[i])if(a[i].v!= Last&AMP;&AMP;!DONE[A[I].V]) {findroot (A[i].v,now); Tree[now].size+=tree[a[i].v].size;if(tree[a[i].v].size>tmp) Tmp=tree[a[i].v].size; }if(sizenow-tree[now].size>tmp) Tmp=sizenow-tree[now].size;if(tmp<size) {root=now; size=tmp;}} void Findroad (intNowint  Last,intRoad) {inti,tmp=0; Len[++t]=road;if(road>m)return; for(i=point[now];i;i=Next[i])if(a[i].v!= Last&AMP;&AMP;!DONE[A[I].V]) findroad (A[I].V,NOW,ROAD+A[I].W);}intCalcintK) {inti,l,r,ret=0;Sort(len+1, len+t+1); for(l=1,r=t;l<r;) {if(len[l]+len[r]<=k) ret+=r-l++;Elser--; }returnRET;} void Work (intNow) {intI sizenow=0; t=0;    Size=inf; Findsize (now,0); Findroot (now,0); Findroad (Root,0,0);    Done[root]=true; Ans+=calc (m); for(i=point[root];i;i=Next[i])if(!DONE[A[I].V]) {t=0; sizenow=0;          Size=inf; Findroad (A[I].V,0,0); Ans-=calc (m-2*a[I].W]; Findsize (A[I].V,0); Findroot (A[I].V,0);            Work (A[I].V); }}intMain () {intIx,y, Z; scanf"%d%d",&n,&m); while(n!=0&&m!=0) {e=0; ans=0; memset (Point,0, sizeof (point)); memset (done,false,sizeof (done)); for(i=1; i<=n-1; ++i) {scanf (" %d%d%d",&x,&y, &z); Addx,y, z); } work (1);printf("%d\ n", ans); scanf"%d%d",&n,&m); }}

A study record of point division in a tree

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