[BZOJ2060][Usaco2010 Nov]Visiting Cows 拜访奶牛

[Usaco2010 Nov]Visiting Cows 拜访奶牛

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

Solution :
树的最大独立集。

Code :

/************************************************************************* > File Name: bzoj2060.cpp > Author: Archer ************************************************************************/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;

const int N = 55555;
struct Node{int v; Node *nxt;}pool[N << 1], *tail = pool, *g[N];
int n, f[N][2];
bool vis[N];
inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') f = 0; ch = getchar();}
    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
    return f ? x : -x;
}
inline void addedge(int u, int v){
    tail->v = v; tail->nxt = g[u]; g[u] = tail++;
    tail->v = u; tail->nxt = g[v]; g[v] = tail++;
}
inline void dfs(int u){
    vis[u] = 1; f[u][0] = 0; f[u][1] = 1;
    for (Node *p = g[u]; p; p = p->nxt) if (!vis[p->v]){
        dfs(p->v); 
        f[u][0] += max(f[p->v][0], f[p->v][1]); f[u][1] += f[p->v][0];
    }
}
int main(){
    n = read();
    for (int i = 1; i < n; i++) addedge(read(), read());
    dfs(1);
    printf("%d\n", max(f[1][0], f[1][1]));
    return 0;
}