You Are Given a Tree
Description
A tree is an undirected graph with exactly one simple path between each pair of vertices. We call a set of simple paths k -valid if each vertex of the tree belongs to no more than one of these paths (including endpoints) and each path consists of exactly k vertices.
You are given a tree with nn vertices. For each k from 1 to nn inclusive find what is the maximum possible size of a k -valid set of simple paths.
Input Format
The first line of the input contains a single integer n ( 2≤n≤100000 ) — the number of vertices in the tree.
Then following n - 1 lines describe the tree, each of them contains two integers v , u ( 1≤v,u≤n ) — endpoints of the corresponding edge.
It is guaranteed, that the given graph is a tree.
Output Format
Output n numbers, the i -th of which is the maximum possible number of paths in an i -valid set of paths.
Sample Input
7
1 2
2 3
3 4
4 5
5 6
6 7Sample Output
7
3
2
1
1
1
1解析
可以先考虑\(k\)确定时的做法，不妨进行树形\(dp\)。\(f[x]\)代表以\(x\)为根的子树中最长链的长度，同时维护一下全局答案。转移方式就是能合并就合并，反之选一条最长的链向上延伸，时间复杂度\(O(n)\)。
我们发现\(k\le\sqrt n\)时答案最多只有\(\sqrt n\)种取值，\(k>\sqrt n\)时答案\(\leq\sqrt n\)，也只有\(\sqrt n\)种取值，并且答案的大小具有单调性，于是就有一个很直观的想法，二分找到段边界，统一每一段的答案即可，时间复杂度\(O(n\sqrt n\log_2n)\)。
但是直接这样写常数可能比较大，换一种整体二分的写法常数更小一些，时间复杂度不变，就可以通过本题了。
\(Code:\)
#include
using namespace std;
const int N = 100020;
struct edge { int ver,next; } e[N*2];
int n,t,Head[N],f[N],cnt,ans[N];
inline void insert(int x,int y) { e[++t] = (edge){y,Head[x]} , Head[x] = t; }
inline void input(void)
{
scanf("%d",&n);
for (int i=1;i= Max ) sec = Max , Max = f[y];
else if ( f[y] > sec ) sec = f[y];
}
if ( sec + Max + 1 >= len ) f[x] = 0 , cnt++;
else f[x] = Max + 1;
}
inline void divide(int st,int ed,int l,int r)
{
if ( st > ed || l > r ) return;
if ( l == r )
{
for (int i=st;i<=ed;i++) ans[i] = l;
return;
}
int mid = st + ed >> 1; cnt = 0;
dp( 1 , 0 , mid );
ans[mid] = cnt;
divide( st , mid-1 , cnt , r );
divide( mid+1 , ed , l , cnt );
}
int main(void)
{
input();
divide( 1 , n , 0 , n );
for (int i=1;i<=n;i++)
printf("%d\n",ans[i]);
return 0;
}