树形dp..水
------------------------------------------------------------------------
#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 MBSubmit: 267 Solved: 198
[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
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}.