"HAOI2009" "P1307" caterpillar

The feeling is much simpler than the other trees, but it's a little bit around (maybe my train of thought is strange.) A) (this is the province of the topic Ah, even as T1 this problem is also too water, ha good weak ... ）

Original question:

For a tree, we can pull a chain and the edge connected to the chain, it looks like a caterpillar, the more points, the bigger the caterpillar. For example, the tree on the left (Figure 1) pulls part of it into a caterpillar on the right (Figure 2).

n≤300000

First get out of the tree without root, this time not enumerate the middle point, say my strange practice one. One

Make two arrays, one is UF, which means the immediate maximum, including yourself, and the other is the BF, which represents the largest of the UF in the brother of X and x down

And then we ask for UF and BF

uf[x]=bf[tree[x].child]+tree[x].cnum;//do not have to reduce UF the biggest son of the inside, because still have their own

If x is a leaf, that is child[x]==0,uf[x]=1

Bf[x]=max (Uf[x],bf[tree[x].brother]);

Because the root is not necessarily in the answer, so set a global maximum ans, after the UF and BF, Ans=max (ans,uf[x]+bf[tree[x].brother]+tree[tree[x].father].cnum-(tree[x). Father==1));

There is no need to subtract two sons, because there are father and master, but when Tree[x].father==1 (I set the root to 1) to 1, because the root has no father

The last direct output of ans can

(Updating the answer with the global maximum should be a lot of DP policies)

Code:

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <cmath>6 using namespacestd;7 intRead () {intz=0, mark=1;CharCh=GetChar ();8      while(ch<'0'|| Ch>'9'){if(ch=='-') mark=-1; Ch=GetChar ();}9      while(ch>='0'&&ch<='9') {z= (z<<3) + (z<<1) +ch-'0'; Ch=GetChar ();}Ten     returnz*Mark; One } A structddd{intNext,y;} e[610000];intlink[310000],ltop=0; -InlinevoidInsertintXintY) {e[++ltop].next=link[x]; link[x]=ltop;e[ltop].y=y;} - structdcd{intBrother,child,father, Cnum;} tree[310000]; theInlinevoidInsert_tree (intXintY) {tree[y].brother=tree[x].child;tree[x].child=y;tree[y].father=x; tree[x].cnum++;} - intn,m; - intuf[310000],bf[310000];//BF represents the largest of many brothers, UF represents the direct maximum - intans=0; + voidGet_tree (intx) { -      for(intI=link[x];i;i=e[i].next)if(e[i].y!=tree[x].father) { + Insert_tree (X,E[I].Y); A Get_tree (E[I].Y); at     } - } - voidDp_tree (intx) { -     if(!x)return ; - Dp_tree (tree[x].brother); -     if(tree[x].child) { in Dp_tree (tree[x].child); -Uf[x]=bf[tree[x].child]+tree[x].cnum;//Do not reduce UF the biggest son of the inside, because there is also his own to     } +     Else -uf[x]=1; thebf[x]=Max (Uf[x],bf[tree[x].brother]); *Ans=max (ans,uf[x]+bf[tree[x].brother]+tree[tree[x].father].cnum-(tree[x].father==1));//not minus two sons, because there's Daddy and grandpa. $ }Panax Notoginseng intMain () {//freopen ("ddd.in", "R", stdin); -memset (UF,0,sizeof(UF)); thememset (BF,0,sizeof(BF)); +cin>>n>>m;//In fact, M is equal to N-1 bar one. A A     int_left,_right; the      for(intI=1; i<=m;i++) {_left=read (), _right=read (); Insert (_left,_right), insert (_right,_left);} +Get_tree (1); -Dp_tree (1); $cout<<ans<<Endl; $     return 0; -}

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"HAOI2009" "P1307" caterpillar