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

2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

Time Limit: 3 Sec  Memory Limit: 64 MB Submit: 474  Solved: 346 [Submit][Status][Discuss]

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}.

 

设$dp[i][0]$表示$i$不选时在以$i$为根的子树中最多能选多少

$dp[i][1]$表示$i$要选时在以$i$为根的子树中最多能选多少

转移显然

#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <algorithm>
using namespace std;
char buf[10000000], *ptr = buf - 1;
inline int readint(){
    int f = 1, n = 0;
    char ch = *++ptr;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = *++ptr;
    }
    while(ch <= '9' && ch >= '0'){
        n = (n << 1) + (n << 3) + ch - '0';
        ch = *++ptr;
    }
    return f * n;
}
const int maxn = 50000 + 10;
struct Edge{
    int to, next;
    Edge(){}
    Edge(int _t, int _n): to(_t), next(_n){}
}e[maxn * 2];
int fir[maxn] = {0}, cnt = 0;
inline void ins(int u, int v){
    e[++cnt] = Edge(v, fir[u]); fir[u] = cnt;
    e[++cnt] = Edge(u, fir[v]); fir[v] = cnt;
}
int dp[maxn][2];
void dfs(int u, int fa){
    dp[u][0] = 0;
    dp[u][1] = 1;
    for(int v, i = fir[u]; i; i = e[i].next){
        v = e[i].to;
        if(v == fa) continue;
        dfs(v, u);
        dp[u][0] += max(dp[v][0], dp[v][1]);
        dp[u][1] += dp[v][0];
    }
}
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    int N = readint();
    for(int u, v, i = 1; i < N; i++){
        u = readint();
        v = readint();
        ins(u, v);
    }
    srand(19260817);
    int root = rand() * rand() % N + 1;
    dfs(root, 0);
    printf("%d
", max(dp[root][0], dp[root][1]));
    return 0;
}