BZOJ 2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛( dp )

树形dp..水

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#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

 

#define rep( i , n ) for( int i = 0 ; i < n ; i++ )

#define clr( x , c ) memset( x , c , sizeof( x ) )

#define REP( x ) for( edge* e = head[ x ] ; e ; e = e -> next )

 

using namespace std;

 

const int maxn = 50000 + 5;

 

struct edge {

int to;

edge* next;

};

 

edge* pt , EDGE[ maxn << 1 ];

edge* head[ maxn ];

 

void edge_init() {

pt = EDGE;

clr( head , 0 );

}

 

void add( int u , int v ) {

pt -> to = v;

pt -> next = head[ u ];

head[ u ] = pt++;

}

#define add_edge( u , v ) add( u , v ) , add( v , u )

 

 

int d[ maxn ][ 2 ];

 

int dp( int x , int k , int fa ) {

int &ans = d[ x ][ k ];

if( ans != -1 )

   return ans;

ans = 0;

if( k ) {

   REP( x ) if( e -> to != fa )

       ans += dp( e -> to , 0 , x );

   ans++;

} else {

REP( x ) if( e -> to != fa ) 

   ans += max( dp( e -> to , 1 , x ) , dp( e -> to , 0 , x ) );

}

return ans;

}

  

int main() {

    freopen( "test.in" , "r" , stdin );

    

    int n;

    cin >> n;

    edge_init();

    rep( i , n - 1 ) {

    int u , v;

    scanf( "%d%d" , &u , &v );

    u-- , v--;

    add_edge( u , v );

    }

    clr( d , -1 );

    cout << max( dp( 0 , 0 , -1 ) , dp( 0 , 1 , -1 ) ) << " ";

return 0;

}

  

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2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 267  Solved: 198
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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