usaco Nov gold 2010 vacation
Problem 3: Visiting Cows [Neal Wu, 2008]
After many weeks of hard work, Bessie is finally getting a vacation!
Being the most social cow in the herd, she wishes to visit her N
(1 <= N <= 50,000) cow friends conveniently 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.
PROBLEM NAME: vacation
INPUT FORMAT:
* Line 1: A single integer: N
* Lines 2..N: Each line describes a single road with two
space-separated integers: C1 and C2
SAMPLE INPUT (file vacation.in):
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
OUTPUT FORMAT:
* Line 1: A single integer representing the maximum number of cows
that Bessie can visit.
SAMPLE OUTPUT (file vacation.out):
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}.
题意是在树中找一些节点，使得这些节点两两不相邻，问满足该条件的点集中节点最多有多少个。
很显然，树状dp ，深搜后对于每一个子树，都有
dp[t][0]+=Max(dp[k][0],dp[k][1]);
dp[t][1]+=dp[k][0];
其中，t 为父节点， k 为子节点， 0 为不选该节点， 1 为选该节点。
代码如下：
#include <stdio.h> #include <vector> using namespace std; #define MAX 50005 vector <int> edge[MAX]; int dp[MAX][2]; int vis[MAX]; int Max(int x,int y) { return x>y?x:y; } void Solve(int t) { int i,j,k; vis[t]=1; dp[t][1]=1; for (i=0;i<edge[t].size();i++) { k=edge[t][i]; if (vis[k]==0) { Solve(k); dp[t][0]+=Max(dp[k][0],dp[k][1]); dp[t][1]+=dp[k][0]; } } } int main() { // freopen("vacation.in","r",stdin); // freopen("vacation.out","w",stdout); int i,j,n,x,y; scanf("%d",&n); for (i=0;i<n;i++) { vis[i]=0; dp[i][0]=dp[i][1]=0; } for (i=0;i<n-1;i++) { scanf("%d%d",&x,&y); x--; y--; edge[x].push_back(y); edge[y].push_back(x); } Solve(0); printf("%d/n",Max(dp[0][0],dp[0][1])); return 0; }