2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛
Time Limit: 3 Sec Memory Limit: 64 MBSubmit: 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
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}.
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
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.