树形dp

https://ac.nowcoder.com/acm/contest/1126/A

链接：https://ac.nowcoder.com/acm/contest/1126/A
来源：牛客网

题目描述

After many weeks of hard work, Bessie is finally getting a vacation! After many weeks of hard work, Bessie is finally getting a vacation! numbered 1..N. The cows have set up quite an unusual road network with exactly N-1 roads connecting pairs of cows C1 and C2 (1 <= C1 <= N; 1 <= C2 <= N; C1 != C2) in such a way that there exists a unique path of roads between any two cows.
FJ wants Bessie to come back to the farm soon; thus, he has instructed Bessie that if two cows are directly connected by a road, she may not visit them both. Of course, Bessie would like her vacation to be as long as possible, so she would like to determine the maximum number of cows she can visit.

输入描述:

* Line 1: A single integer: N
* Lines 2..N: Each line describes a single road with two space-separated integers: C1 and C2

输出描述:

* Line 1: A single integer representing the maximum number of cows that Bessie can visit.

示例1

输入

复制

7 
6 2 
3 4 
2 3 
1 2 
7 6 
5 6 

输出

复制

4

说明

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.
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}.

https://www.cnblogs.com/seaupnice/p/9471700.html

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
const int Maxn = 50000+5;

vector<int> G[Maxn];
int vi[Maxn] , dp[Maxn][2];

int dfs(int u)
{
    dp[u][1] = 1 ;
    for(int i = 0 ; i < G[u].size() ; i++)
    {
        int v = G[u][i];
        if(!vi[v])
        {
            vi[v] = 1 ;
            dfs(v);
            dp[u][0] += max(dp[v][0] , dp[v][1]);
            dp[u][1] += dp[v][0];
        }
    }
}

int main()
{
    int n ;
    scanf("%d" , &n);
    for(int i = 1 ; i < n ; i++)
    {
        int x , y ;
        scanf("%d%d" , &x , &y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    memset(vi , 0 , sizeof(vi));
    memset(dp , 0 , sizeof(dp));
    vi[1] = 1 ;
    dfs(1);
    cout << max(dp[1][0] , dp[1][1]) << endl ;

    return 0 ;
}