2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 252  Solved: 185
[Submit][Status]

Description

经过了几周的辛苦工作,贝茜终于迎来了一个假期.作为奶牛群中最会社交的牛,她希望去拜访N(1<=N<=50000)个朋友.这些朋友被标号为1..N.这些奶牛有一个不同寻常的交通系统,里面有N-1条路,每条路连接了一对编号为C1和C2的奶牛(1 <= C1 <= N; 1 <= C2 <= N; C1<>C2).这样,在每一对奶牛之间都有一条唯一的通路. FJ希望贝茜尽快的回到农场.于是,他就指示贝茜,如果对于一条路直接相连的两个奶牛,贝茜只能拜访其中的一个.当然,贝茜希望她的假期越长越好,所以她想知道她可以拜访的奶牛的最大数目.

Input

第1行:单独的一个整数N 第2..N行:每一行两个整数,代表了一条路的C1和C2.

Output

单独的一个整数,代表了贝茜可以拜访的奶牛的最大数目.

Sample Input

7
6 2
3 4
2 3
1 2
7 6
5 6

INPUT DETAILS:

Bessie knows 7 cows. Cows 6 and 2 are directly connected by a road,
as are cows 3 and 4, cows 2 and 3, etc. The illustration below depicts the
roads that connect the cows:

1--2--3--4
|
5--6--7

Sample Output

4

OUTPUT DETAILS:

Bessie can visit four cows. The best combinations include two cows
on the top row and two on the bottom. She can't visit cow 6 since
that would preclude visiting cows 5 and 7; thus she visits 5 and
7. She can also visit two cows on the top row: {1,3}, {1,4}, or
{2,4}.

HINT

 

Source

Gold

题解：树状DP啦啦树状DP，按照下面的子节点选和不选进行转移完事

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 type
     point=^node;
     node=record
                g:longint;
                next:point;
     end;
     vec=record
               a0,a1:longint;
     end;
 var
    i,j,k,l,m,n:longint;
    a:array[..] of point;
    b:array[..] of longint;
    t:vec;
 function max(x,y:longint):longint;inline;
          begin
               if x>y then max:=x else max:=y;
          end;
 procedure add(x,y:longint);inline;
           var p:point;
           begin
                new(p);p^.g:=y;
                p^.next:=a[x];a[x]:=p;
           end;
 function dfs(x:longint):vec;inline;
          var p:point;t,v:vec;
          begin
               b[x]:=;
               p:=a[x];
               t.a0:=;t.a1:=;
               while p<>nil do
                     begin
                          if b[p^.g]= then
                             begin
                                  v:=dfs(p^.g);
                                  t.a1:=t.a1+v.a0;
                                  t.a0:=t.a0+max(v.a1,v.a0);
                             end;
                          p:=p^.next;
                     end;
               exit(t);
          end;
 begin
      readln(n);
      for i:= to n do a[i]:=nil;
      fillchar(b,sizeof(b),);
      for i:= to n- do
          begin
               readln(j,k);
               add(j,k);add(k,j);
          end;
      t:=dfs();
      writeln(max(t.a0,t.a1));
 end.