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}.
题解：f[i][0]为不选根节点时最大人数 f[i][1]为选择根节点时最大人数
大力dp一波
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn=50000+10;
vector<int>A[maxn];
int f[maxn][2];
inline void dfs(int x,int fa){f[x][0]=0;f[x][1]=1;for(int i=0;i<A[x].size();i++){int u=A[x][i];if(u==fa)continue;dfs(u,x);f[x][0]+=max(f[u][0],f[u][1]);f[x][1]+=f[u][0];}
}
int main(){//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);int n;scanf("%d",&n);int x,y;for(int i=1;i<n;i++){scanf("%d %d",&x,&y);A[x].push_back(y);A[y].push_back(x);}dfs(1,0);printf("%d\n",max(f[1][0],f[1][1]));
return 0;
}