Time Limit: 3 Sec Memory Limit: 64 MB
Submit: 252 Solved: 185
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.
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.
单独的一个整数,代表了贝茜可以拜访的奶牛的最大数目.
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
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}.
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
1 /**************************************************************
2 Problem: 2060
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:112 ms
7 Memory:2576 kb
8 ****************************************************************/
9
10 type
11 point=^node;
12 node=record
13 g:longint;
14 next:point;
15 end;
16 vec=record
17 a0,a1:longint;
18 end;
19 var
20 i,j,k,l,m,n:longint;
21 a:array[0..100000] of point;
22 b:array[0..100000] of longint;
23 t:vec;
24 function max(x,y:longint):longint;inline;
25 begin
26 if x>y then max:=x else max:=y;
27 end;
28 procedure add(x,y:longint);inline;
29 var p:point;
30 begin
31 new(p);p^.g:=y;
32 p^.next:=a[x];a[x]:=p;
33 end;
34 function dfs(x:longint):vec;inline;
35 var p:point;t,v:vec;
36 begin
37 b[x]:=1;
38 p:=a[x];
39 t.a0:=0;t.a1:=1;
40 while p<>nil do
41 begin
42 if b[p^.g]=0 then
43 begin
44 v:=dfs(p^.g);
45 t.a1:=t.a1+v.a0;
46 t.a0:=t.a0+max(v.a1,v.a0);
47 end;
48 p:=p^.next;
49 end;
50 exit(t);
51 end;
52 begin
53 readln(n);
54 for i:=1 to n do a[i]:=nil;
55 fillchar(b,sizeof(b),0);
56 for i:=1 to n-1 do
57 begin
58 readln(j,k);
59 add(j,k);add(k,j);
60 end;
61 t:=dfs(1);
62 writeln(max(t.a0,t.a1));
63 end.