comet-2-D(dfs+容斥)

题目链接

https://www.cometoj.com/contest/37/problem/D?problem_id=1531

题解

需要提取每个联通块的特征,再依次枚举,这个特征就是直径的中点了,显然直径的中点必然相同,那么固定中点去计数就可以了

中点可以在点上也可以在边上,然后 $DFS$ 统计不同深度的点数的前缀和,那么深度至多为 $k$ 的点集为 $2^{num_k}$ 个,然后做一下容斥就可以了。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 2005
#define nm 4005
using namespace std;
const double pi=acos(-1);
const ll inf=998244353;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}




struct edge{int t;edge*next;}e[nm],*h[NM],*o=e;
void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;}
int n,_x,_y,d[NM][NM];
ll p[NM],ans[NM];

void dfs(int x,int f,int t){
d[_x][t]++;
link(x)if(j->t!=f)dfs(j->t,x,t+1);
}


int main(){
n=read();inc(i,2,n){_x=read();_y=read();add(_x,_y);add(_y,_x);}
p[0]=1;inc(i,1,n)p[i]=p[i-1]*2%inf;
ans[1]=n-1;
inc(i,1,n)link(i)if(i<j->t){
mem(d[i]);
dfs(_x=i,j->t,0);
inc(k,1,n)d[_x][k]+=d[_x][k-1];
mem(d[j->t]);
dfs(_x=j->t,i,0);
inc(k,1,n)d[_x][k]+=d[_x][k-1];
inc(k,1,n/2){
ans[k<<1|1]+=p[d[i][k]]*p[d[j->t][k]]%inf;
ans[k<<1|1]-=p[d[i][k-1]]*p[d[j->t][k-1]]%inf;
ans[k<<1|1]-=(p[d[i][k]]-p[d[i][k-1]])*p[d[j->t][k-1]]%inf;
ans[k<<1|1]-=p[d[i][k-1]]*(p[d[j->t][k]]-p[d[j->t][k-1]])%inf;
ans[k<<1|1]%=inf;
}
}
inc(i,1,n){
link(i){
mem(d[j->t]);
dfs(_x=j->t,i,1);
inc(k,1,n)d[_x][k]+=d[_x][k-1];
}
mem(d[i]);
dfs(_x=i,i,0);
inc(k,1,n)d[_x][k]+=d[_x][k-1];
inc(k,1,n/2){
ans[k<<1]+=p[d[i][k]]-p[d[i][k-1]],ans[k<<1]%=inf;
link(i){
ans[k<<1]-=p[d[i][k-1]-d[j->t][k-1]]*(p[d[j->t][k]]-p[d[j->t][k-1]])%inf;
ans[k<<1]%=inf;
}
}
}
inc(i,1,n-1)printf("%lld\n",(ans[i]+inf)%inf);
return 0;
}