bzoj 2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛(树形DP)

2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

Time Limit: 3 Sec Memory Limit: 64 MB

Submit: 346 Solved: 253

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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}.

HINT

Source

Gold

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 50003
using namespace std;
int n,m,next[N*2],point
,u[N*2],tot,f
[3];
void add(int x,int y)
{
tot++; next[tot]=point[x]; point[x]=tot; u[tot]=y;
tot++; next[tot]=point[y]; point[y]=tot; u[tot]=x;
}
void dfs(int x,int fa)
{
for (int i=point[x];i;i=next[i])
if (u[i]!=fa)
{
dfs(u[i],x);
f[x][0]=f[x][0]+max(f[u[i]][0],f[u[i]][1]);
f[x][1]=f[x][1]+f[u[i]][0];
}
f[x][1]++;
}
int main()
{
scanf("%d",&n);
for (int i=1;i<n;i++)
{
int x,y; if (x>y) swap(x,y);
scanf("%d%d",&x,&y);
add(x,y);
}
dfs(1,0);
printf("%d",max(f[1][1],f[1][0]));
return 0;
}