bzoj 2060（树形DP）

2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

Time Limit: 3 Sec   Memory Limit: 64 MB
Submit: 335   Solved: 244
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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}.

解题思路：简单的树形DP。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int n,len;
int to[100005],next[100005];
int h[100005];
int f[2][100005];


inline int read()
{
char y;int x=0,f=1; y=getchar();
while (y<'0' || y>'9') {if (y=='-') f=-1; y=getchar();}
while (y>='0' && y<='9') {x=x*10+int(y)-48; y=getchar();}
return x*f;
}


void insert(int x,int y)
 {
  ++len; to[len]=y; next[len]=h[x]; h[x]=len;
 }


void dfs(int o,int fa)
 {
  f[1][o]=1; f[0][o]=0;
  int u=h[o];
  while (u!=0)
  {
   if (to[u]!=fa)
    {
    dfs(to[u],o);
    f[1][o]+=f[0][to[u]];
    f[0][o]+=max(f[1][to[u]],f[0][to[u]]);
  }
 u=next[u];
}
 }


int main()
{
n=read();
for (int i=1;i<=n-1;++i)
{
int x,y; x=read(); y=read();
insert(x,y); insert(y,x);
}
memset(f,-0x7f,sizeof(f));
dfs(1,0);
printf("%d",max(f[1][1],f[0][1]));
}