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AC的做法

枚举颜色，然后考虑我可以点亮某个点或者熄灭某个点，以点亮某个点为例，假设这个点是iii，枚举树上已经被点亮的点jjj，那么此时答案的增量为
∑j(depthi+depthj−2depthlca(i,j))wiwjP\sum_{j} (depth_i+depth_j-2depth_{lca(i,j)})w_iw_jP j∑(depthi+depthj−2depthlca(i,j))wiwjP

其中P=∏(ri−li+1),wi=(ri−li+1)−1P = \prod (r_i-l_i+1),w_i = (r_i-l_i+1)^{-1}P=∏(ri−li+1),wi=(ri−li+1)−1

化简一下得到
Pwi(depthi∑jdepthj+∑jwjdepthj−2∑jdepthlca(i,j)wj)P w_i \left( depth_i \sum_{j} depth_j + \sum_{j} w_j depth_j - 2 \sum_{j} depth_{lca(i,j)}w_j \right) Pwi(depthij∑depthj+j∑wjdepthj−2j∑depthlca(i,j)wj)

一共三项，前两项都很容易维护

第三项，我可以对于已存在的jjj，把根节点到jjj的每个节点的权值都+=wj+=w_j+=wj，那么iii到根节点这条路径上的权值之和就是第三项

AC的代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 100010
#define maxe 200010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
#define mod 1000000007ll
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{ll c, f(1);for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;for(;isdigit(c);c=getchar())x=x*10+c-0x30;return f*x;
}
ll ans, n, l[maxn], r[maxn], inv[maxn], P, now, w[maxn];
struct SegmentTree
{ll sum[maxn<<2], add[maxn<<2], set[maxn<<2], L[maxn<<2], R[maxn<<2];void maketag_set(ll o, ll v){add[o]=0;set[o]=v;sum[o]=(R[o]-L[o]+1)*v;}void maketag_add(ll o, ll v){(add[o]+=v)%=mod;(sum[o]+=(R[o]-L[o]+1)*v)%=mod;}void pushdown(ll o){if(L[o]==R[o])return;if(~set[o]){maketag_set(o<<1,set[o]);maketag_set(o<<1|1,set[o]);set[o]=-1;}if(add[o]){maketag_add(o<<1,add[o]);maketag_add(o<<1|1,add[o]);add[o]=0;}}void pushup(ll o){(sum[o]=sum[o<<1]+sum[o<<1|1])%mod;}void build(ll o, ll l, ll r){ll mid(l+r>>1);L[o]=l, R[o]=r;add[o]=0;set[o]=-1;if(l==r){return;}build(o<<1,l,mid);build(o<<1|1,mid+1,r);pushup(o);}void segset(ll o, ll l, ll r, ll v){ll mid(L[o]+R[o]>>1);if(l<=L[o] and r>=R[o]){maketag_set(o,v);return;}pushdown(o);if(l<=mid)segset(o<<1,l,r,v);if(r>mid)segset(o<<1|1,l,r,v);pushup(o);}void segadd(ll o, ll l, ll r, ll v){ll mid(L[o]+R[o]>>1);if(l<=L[o] and r>=R[o]){maketag_add(o,v);return;}pushdown(o);if(l<=mid)segadd(o<<1,l,r,v);if(r>mid)segadd(o<<1|1,l,r,v);pushup(o);}ll segsum(ll o, ll l, ll r){pushdown(o);ll mid(L[o]+R[o]>>1), ans(0);if(l<=L[o] and r>=R[o])return sum[o];if(l<=mid)ans+=segsum(o<<1,l,r);if(r>mid)ans+=segsum(o<<1|1,l,r);return ans%mod;}
}segtree;
struct Graph
{int etot, head[maxn], to[maxe], next[maxe], w[maxe];void clear(int N){for(int i=1;i<=N;i++)head[i]=0;etot=0;}void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}#define forp(pos,G) for(auto p=G.head[pos];p;p=G.next[p])
}G;
struct Heavy_Light_Decomposition
{int size[maxn], top[maxn], tid[maxn], tim, untid[maxn], depth[maxn], son[maxn], fa[maxn];void dfs1(Graph &G, int pos){int p, v;size[pos]=1;for(p=G.head[pos];p;p=G.next[p]){if((v=G.to[p])==fa[pos])continue;fa[v]=pos;depth[v]=depth[pos]+1;dfs1(G,v);if(size[v]>size[son[pos]])son[pos]=v;size[pos]+=size[v];}}void dfs2(Graph &G, int pos, int tp){int p, v;top[pos]=tp;tid[pos]=++tim;untid[tid[pos]]=pos;if(son[pos])dfs2(G,son[pos],tp);for(p=G.head[pos];p;p=G.next[p])if((v=G.to[p])!=fa[pos] and v!=son[pos])dfs2(G,v,v);}void run(Graph &G, int root){tim=0;depth[root]=1;dfs1(G,root);dfs2(G,root,root);}
}HLD;
vector<pll> events[maxn];
int main()
{ll n, i;inv[1]=1;for(int i=2;i<maxn;i++)inv[i]=inv[mod%i]*(mod-mod/i)%mod;n=read();P=1;rep(i,n){l[i]=read(), r[i]=read();(P*=r[i]-l[i]+1)%=mod;w[i]=inv[r[i]-l[i]+1];events[l[i]].emb( pll(i,+1) );events[r[i]+1].emb( pll(i,-1) );}rep(i,n-1){int u=read(), v=read();G.adde(u,v), G.adde(v,u);}HLD.run(G,1);segtree.build(1,1,n);ll sw=0, sdw=0;rep(i,1e5){for(auto e:events[i]){auto pos=e.fi, v=e.se;ll delt=0;(delt += HLD.depth[pos]*sw)%=mod;(delt += sdw)%=mod;ll s=0, x=pos;while(x){auto tp = HLD.top[x];(s+=segtree.segsum(1,HLD.tid[tp],HLD.tid[x]))%=mod;x=HLD.fa[tp];}(delt -= 2*s)%=mod;(now+=v*P*w[pos]%mod*delt)%=mod;x=pos;while(x){auto tp = HLD.top[x];segtree.segadd(1,HLD.tid[tp],HLD.tid[x],w[pos]*v);x=HLD.fa[tp];}(sw+=v*w[pos])%=mod;(sdw+=v*w[pos]*HLD.depth[pos])%=mod;}(ans+=now)%=mod;}cout<<(ans+mod)%mod;return 0;
}

因为常数太大而TLETLETLE的做法

点分治，一棵子树一棵子树的进行合并

新来一棵子树的时候，我就对于每个节点都求一下它颜色区间的深度和，这是路径的一段，另一段可以通过维护每种颜色被多少个点包含了进而算出来

两个线段树可能常数比较大？
而且实际上线段树操作的调用次数也比前一种算法多，所以就TLETLETLE了

因为常数太大而TLETLETLE的代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 100010
#define maxe 200010
#define maxk 17
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
#define mod 1000000007ll
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{ll c, f(1);for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;for(;isdigit(c);c=getchar())x=x*10+c-0x30;return f*x;
}
ll ans, n, l[maxn], r[maxn], inv[maxn], P;
struct SegmentTree
{ll sum[maxn<<2], add[maxn<<2], set[maxn<<2], L[maxn<<2], R[maxn<<2];void maketag_set(ll o, ll v){add[o]=0;set[o]=v;sum[o]=(R[o]-L[o]+1)*v;}void maketag_add(ll o, ll v){(add[o]+=v)%=mod;(sum[o]+=(R[o]-L[o]+1)*v)%=mod;}void pushdown(ll o){if(L[o]==R[o])return;if(~set[o]){maketag_set(o<<1,set[o]);maketag_set(o<<1|1,set[o]);set[o]=-1;}if(add[o]){maketag_add(o<<1,add[o]);maketag_add(o<<1|1,add[o]);add[o]=0;}}void pushup(ll o){(sum[o]=sum[o<<1]+sum[o<<1|1])%mod;}void build(ll o, ll l, ll r){ll mid(l+r>>1);L[o]=l, R[o]=r;add[o]=0;set[o]=-1;if(l==r){return;}build(o<<1,l,mid);build(o<<1|1,mid+1,r);pushup(o);}void segset(ll o, ll l, ll r, ll v){ll mid(L[o]+R[o]>>1);if(l<=L[o] and r>=R[o]){maketag_set(o,v);return;}pushdown(o);if(l<=mid)segset(o<<1,l,r,v);if(r>mid)segset(o<<1|1,l,r,v);pushup(o);}void segadd(ll o, ll l, ll r, ll v){ll mid(L[o]+R[o]>>1);if(l<=L[o] and r>=R[o]){maketag_add(o,v);return;}pushdown(o);if(l<=mid)segadd(o<<1,l,r,v);if(r>mid)segadd(o<<1|1,l,r,v);pushup(o);}ll segsum(ll o, ll l, ll r){pushdown(o);ll mid(L[o]+R[o]>>1), ans(0);if(l<=L[o] and r>=R[o])return sum[o];if(l<=mid)ans+=segsum(o<<1,l,r);if(r>mid)ans+=segsum(o<<1|1,l,r);return ans%mod;}
}sum_tree, cnt_tree;
struct Graph
{int etot, head[maxn], to[maxe], next[maxe], w[maxe];void clear(int N){for(int i=1;i<=N;i++)head[i]=0;etot=0;}void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}#define forp(pos,G) for(auto p=G.head[pos];p;p=G.next[p])
}G;
struct Centroid_Decomposition
{ll grey[maxn], depth[maxn], size[maxn], dist[maxn], lis[maxn];void dfs(ll pos, ll pre, Graph& G){lis[++*lis]=pos;// lis.emb(pos);// de(pos);size[pos]=1;forp(pos,G)if(G.to[p]!=pre and !grey[G.to[p]]){depth[G.to[p]]=depth[pos]+1;dist[G.to[p]]=dist[pos]+G.w[p];dfs(G.to[p],pos,G);size[pos]+=size[G.to[p]];}}void findG(ll pos, ll pre, Graph& G, ll& g, ll& gsum, ll nowsum, ll node_tot){if(nowsum<gsum)g=pos, gsum=nowsum;forp(pos,G)if(G.to[p]!=pre and !grey[G.to[p]])findG( G.to[p], pos, G, g, gsum, nowsum+(node_tot-2*size[G.to[p]]), node_tot );}void solve(ll pos, Graph& G){*lis=0; dfs(pos,0,G);ll g=pos, gsum=0, i;rep(i,*lis)gsum+=depth[lis[i]];//, cerr<<x<<'('<<size[x]<<')'<<' ';cerr<<endl;findG(pos,0,G,g,gsum,gsum,*lis);grey[g]=1;//--- process ---sum_tree.segset(1,1,1e5,0);cnt_tree.segset(1,1,1e5,0);forp(g,G)if(!grey[G.to[p]]){depth[G.to[p]]=1;*lis=0;dfs(G.to[p],g,G);rep(i,*lis){auto x=lis[i];ans += sum_tree.segsum(1,l[x],r[x])*inv[r[x]-l[x]+1]%mod;ans += cnt_tree.segsum(1,l[x],r[x])*depth[x]%mod*inv[r[x]-l[x]+1]%mod;ans%=mod;}rep(i,*lis){auto x=lis[i];sum_tree.segadd(1,l[x],r[x],depth[x]*inv[r[x]-l[x]+1]%mod*P%mod);cnt_tree.segadd(1,l[x],r[x],inv[r[x]-l[x]+1]*P%mod);}}(ans+=sum_tree.segsum(1,l[g],r[g])*inv[r[g]-l[g]+1])%=mod;   //不要忘记处理一端在重心的路径//---------------forp(g,G)if(!grey[G.to[p]])solve(G.to[p],G);}
}CD;
int main()
{int n, i;inv[1]=1;for(int i=2;i<maxn;i++)inv[i]=inv[mod%i]*(mod-mod/i)%mod;n=read();P=1;rep(i,n){l[i]=read(), r[i]=read();(P*=r[i]-l[i]+1)%=mod;}rep(i,n-1){int u=read(), v=read();G.adde(u,v), G.adde(v,u);}// de("debug1");sum_tree.build(1,1,1e5);cnt_tree.build(1,1,1e5);// de("debug2");CD.solve(1,G);cout<<ans;return 0;
}

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cf1260 F. Colored Tree

链接

AC的做法

AC的代码

因为常数太大而TLETLETLE的做法

因为常数太大而TLETLETLE的代码

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