Portal
Solution
Leaf nodes change interval is continuous, that weight change interval non-leaf nodes is also continuous
It can be seen, \ (W is \) of the feasible region is continuous change in value, so it can simply look becomes \ (W + 1 \) or \ (W-1 \)
For the answer, you can ask the energy consumption of no more than \ (k \) is a set of numbers \ (ans_k \) , then check points
Found \ (ans_1 = 2 ^ {m -1}, ans_n = 2 ^ {m} -1 \)
First discovered, if you want (W \) \ value \ (+ 1 \) , then only to touch those \ (\ leq W \) of the node, so that their weight increases
Similarly, we found no cross process two types of nodes, it can be determined that each \ (W is \) plus one or a minus probability
Because the point by reducing the weight of a large weight can be substituted approach above method
Iff contains \ (W is \) point, it takes only \ (1 \) energy
Next consider not contain \ (W \) case point, if we ask \ (ans_k \)
The \ (<W \) of small nodes are called node, \ (> W is \) node is referred to as large nodes
Set \ (f_i \) indicates that only small changes in the value of the node but \ (i \) point weights still \ (\ leq W \) the probability of
Odd dot:
\ [f_u = \ Prod f_v \]
even dots:
\ [l- f_u = \ Prod (-f_v. 1) \]
provided \ (G_i \) represents may be such that by changing the large nodes \ (I \) point weight \ (<W \) probability of metastasis (f_i \) \ similarTherefore, to obtain
\ [ans_k = 2 ^ {m
-1} (1-f_1 (1-g_1)) + 2 ^ {m-1} \] where \ (f_1 (1-g_1) \) represents the root probability weights remain unchanged\ (k \) increments every \ (1 \) Up to two leaves change points \ (dp \) value
Dynamic \ (dp \) to achieve
For leaf depth of the parity issue
Double layer node may enable \ (f_i '= 1-f_i , g_i' = 1-g_i \)
In this way, we can be shifted by this
\ [f_u = \ prod (1
-f_v) \\ g_u = \ prod (1-g_v) \] is the cancer problem. . .有可能会除以\(0\)，所以记录一下乘了\(0\)的个数，和不含\(0\)的其它项的积
突然发现自己忘记ddp的做法，不妨重温一下
首先我们设了函数\(f\)和\(g\)，(和上述函数无关）
分别表示算了重儿子和没算重儿子的答案
在本题中，它们的转移如下
如果是叶子节点，或者某个重链的链底
则有\(g_i=f_i\)
然后有（序号表示的是距离链头的距离\(+1\)）
\[ f_1=(1-f_2)g_1 \\f_2=(1-f3)g_2 \\...\\f_n=g_n\\ f_1=g_1-g_1g_2+g_1g_3g_4-...+(-1)^{n+1}(g_1g_2...g_n) \]
由此可以发现，在维护那个线段树区间乘法的同时，还需要维护一个如上的\(Sum\)值总结：
之前做ddp的时候，是使用线段树维护一个链的转移矩阵的积，并不能直接求出链顶的答案
而在此题，我们直接将修改链底的答案，并更新进线段树，通过计算直接得到链顶的答案，线段树可以直接取代\(f\)数组
另外不需要建立庞大的所有点的线段树
可以分别对每条链建一棵，这样就只需要一个\(Modify\)，而不用\(Query\)
最后，这真是一道毒瘤题
Code
#include<bits/stdc++.h>
#define ll long long
using namespace std;
#define reg register
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
return x*f;
}
const int P=998244353,MN=2e5+5;
int Mul(int x,int y){return (1ll*x*y)%P;}
int Add(int x,int y){return (x+y)%P;}
int fpow(int x,int m){int r=1;for(;m;m>>=1,x=Mul(x,x))if(m&1)r=Mul(r,x);return r;}
struct node
{
int a,cnt;
node(){a=1,cnt=0;}
node(int a,int cnt):a(a),cnt(cnt){}
void mul(node o){a=Mul(a,o.a),cnt+=o.cnt;}
void div(node o){a=Mul(a,fpow(o.a,P-2)),cnt-=o.cnt;}
int val(){return cnt?0:a;}
}f[MN],g[MN];
node _(int x){x%=P;return x?node(x,0):node(1,1);}
struct edge{int to,nex;}e[MN<<1];
bool lf[MN];
int hr[MN],en,mx[MN],siz[MN],top[MN],bot[MN],fa[MN],dfn[MN],fdfn[MN],dep[MN],dind=0,leaf=1;
int n,W,ans[MN],F[MN],G[MN];
inline void ins(int x,int y)
{
e[++en]=(edge){y,hr[x]};hr[x]=en;
e[++en]=(edge){x,hr[y]};hr[y]=en;
}
void rw(int &x,int y,int z){z?x=max(x,y):x=min(x,y);}
int dfs1(int x,int f,int d)
{
register int i;fa[x]=f;siz[x]=1;dep[x]=d;
int res=d&1?0:0x3f3f3f3f;
for(i=hr[x];i;i=e[i].nex)if(e[i].to^f)
{
rw(res,dfs1(e[i].to,x,d+1),d&1);
siz[x]+=siz[e[i].to],siz[e[i].to]>siz[mx[x]]?mx[x]=e[i].to:0;
}
if(siz[x]==1){lf[x]=1;leaf=(leaf<<1)%P;return x;}
return res;
}
int rt[MN],ls[MN<<2],rs[MN<<2],sum_f[MN<<2],sum_g[MN<<2],prd_f[MN<<2],prd_g[MN<<2],tt;
void Build(int &x,int l,int r);
void dfs2(int x,int tp)
{
fdfn[dfn[x]=++dind]=x;top[x]=tp;
register int i;if(mx[x])dfs2(mx[x],tp);
for(i=hr[x];i;i=e[i].nex)if((e[i].to^fa[x])&&(e[i].to^mx[x]))dfs2(e[i].to,e[i].to);
if(mx[x])
{
F[x]=G[x]=1;
for(i=hr[x];i;i=e[i].nex)if(e[i].to^fa[x])
{
F[x]=Mul(F[x],(1+P-F[e[i].to]));
G[x]=Mul(G[x],(1+P-G[e[i].to]));
if(e[i].to^mx[x])
f[x].mul(_(1+P-F[e[i].to])),g[x].mul(_(1+P-G[e[i].to]));
}
}
else
{
F[x]=(x>W)^(dep[x]&1),G[x]=(x>=W)^(dep[x]&1);
bot[top[x]]=x;
f[x]=_((x>W)^(dep[x]&1));g[x]=_((x>=W)^(dep[x]&1));
}
if(x==tp) Build(rt[x],dfn[x],dfn[bot[x]]);
}
void up(int x,int l)
{
sum_f[x]=Add(sum_f[ls[x]],Mul((l&1?P-prd_f[ls[x]]:prd_f[ls[x]]),sum_f[rs[x]]));
prd_f[x]=Mul(prd_f[ls[x]],prd_f[rs[x]]);
sum_g[x]=Add(sum_g[ls[x]],Mul((l&1?P-prd_g[ls[x]]:prd_g[ls[x]]),sum_g[rs[x]]));
prd_g[x]=Mul(prd_g[ls[x]],prd_g[rs[x]]);
}
void update(int x,int a){sum_f[x]=prd_f[x]=f[a].val();sum_g[x]=prd_g[x]=g[a].val();}
void Build(int &x,int l,int r)
{
x=++tt;
if(l==r)return(void)(update(x,fdfn[l]));
int mid=(l+r)>>1;
Build(ls[x],l,mid);Build(rs[x],mid+1,r);
up(x,mid-l+1);
}
void Modify(int x,int l,int r,int a)
{
if(l==r)return(void)(update(x,fdfn[l]));
int mid=(l+r)>>1;
a<=mid?Modify(ls[x],l,mid,a):Modify(rs[x],mid+1,r,a);
up(x,mid-l+1);
}
#define o top[x]
void solve_f(int x)
{
if(fa[o]) f[fa[o]].div(_(P+1-sum_f[rt[o]]));
Modify(rt[o],dfn[o],dfn[bot[o]],dfn[x]);
if(fa[o]) f[fa[o]].mul(_(P+1-sum_f[rt[o]])),solve_f(fa[o]);
}
void solve_g(int x)
{
if(fa[o]) g[fa[o]].div(_(P+1-sum_g[rt[o]]));
Modify(rt[o],dfn[o],dfn[bot[o]],dfn[x]);
if(fa[o]) g[fa[o]].mul(_(P+1-sum_g[rt[o]])),solve_g(fa[o]);
}
int main()
{
int L,R;
n=read();L=read();R=read();
register int i,j;
for(i=1;i<n;++i) j=read(),ins(j,read());
W=dfs1(1,0,1);dfs2(1,1);
ans[1]=leaf=Mul(leaf,(P+1)>>1);
for(i=2;i<n;++i)
{
if(W+1-i>=1&&lf[W+1-i]) f[W+1-i]=_(P+1>>1),solve_f(W+1-i);
if(W+i-1<=n&&lf[W+i-1]) g[W+i-1]=_(P+1>>1),solve_g(W+i-1);
ans[i]=Mul(Add(Mul(P-sum_f[rt[1]],P+1-sum_g[rt[1]]),2),leaf);
}
ans[n]=Add(Mul(leaf,2),P-1);
for(i=L;i<=R;++i)
printf("%d ",Add(ans[i],P-ans[i-1]));
return 0;
}Blog来自PaperCloud，未经允许，请勿转载，TKS！