神题!
虽然想到离线,但是还是只会随机数据的做法。。。。
有一个比较有意思的结论:把点i到根的所有点权值设为1,其他点为0,此时j到根的所有点权和即为dep[LCA(i,j)]
不难发现,这个方法满足加法性质。
把询问查分,从1到n处理每个点,将其到根的权值+1,询问只要看这个点到根的权值和就是LCA的深度和了!
LCT维护即可
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long LL; const int Maxn = 50005; const int Mod = 201314; int sum[Maxn], size[Maxn]; int fa[Maxn], rev[Maxn]; int w[Maxn], tip[Maxn]; int son[Maxn][2], ans[Maxn]; int n,m,x,l,r,i,j; struct arr { int num,x,ans; bool operator <(const arr &a)const { return num<a.num; } } q[Maxn*2]; #define isroot(x) (son[fa[(x)]][0]!=(x)&&son[fa[(x)]][1]!=(x)) void update(int x){ (sum[x] = sum[son[x][0]] + sum[son[x][1]] + w[x]) %= Mod; size[x] = size[son[x][0]] + size[son[x][1]] + 1; } void push1(int x){ swap(son[x][0], son[x][1]); rev[son[x][0]] ^= 1; rev[son[x][1]] ^= 1; rev[x] = 0; } void push2(int x){ if (son[x][0]){ (w[son[x][0]] += tip[x]) %= Mod; (sum[son[x][0]] += (LL)tip[x]*size[son[x][0]]%Mod) %= Mod; (tip[son[x][0]] += tip[x]) %= Mod; } if (son[x][1]){ (w[son[x][1]] += tip[x]) %= Mod; (sum[son[x][1]] += (LL)tip[x]*size[son[x][1]]%Mod) %= Mod; (tip[son[x][1]] += tip[x]) %= Mod; } tip[x] = 0; } void rotate(int x,int ft,int K){ if (!isroot(ft)){ if (son[fa[ft]][0]==ft) son[fa[ft]][0]=x; else son[fa[ft]][1]=x; } fa[x] = fa[ft]; if (son[x][K^1]) fa[son[x][K^1]]=ft; son[ft][K] = son[x][K^1]; fa[ft]=x; son[x][K^1]=ft; update(ft); } void Splay(int x){ int ft, gf; while (!isroot(x)){ ft = fa[x]; gf = fa[ft]; if (rev[gf]) push1(gf); if (tip[gf]) push2(gf); if (rev[ft]) push1(ft); if (tip[ft]) push2(ft); if (rev[x]) push1(x); if (tip[x]) push2(x); if (isroot(ft)){ if (son[ft][0]==x) rotate(x,ft,0); if (son[ft][1]==x) rotate(x,ft,1); } else { if (son[ft][0]==x && son[gf][0]==ft) rotate(ft,gf,0), rotate(x,ft,0); if (son[ft][1]==x && son[gf][1]==ft) rotate(ft,gf,1), rotate(x,ft,1); if (son[ft][0]==x && son[gf][1]==ft) rotate(x,ft,0), rotate(x,gf,1); if (son[ft][1]==x && son[gf][0]==ft) rotate(x,ft,1), rotate(x,gf,0); } } if (rev[x]) push1(x); if (tip[x]) push2(x); update(x); } void access(int x){ int t = 0; while (x){ Splay(x); son[x][1] = t; update(x); t = x; x = fa[x]; } } void makeroot(int x) { access(x); Splay(x); rev[x]^=1; } void Link(int x,int y) { makeroot(x); access(y); fa[x]=y; } void Cut(int x,int y){ makeroot(x); access(y); Splay(y); son[y][0] = 0; fa[x] = 0; Splay(x); Splay(y); } int main(){ //freopen("3626.in","r",stdin); //freopen("3626.out","w",stdout); scanf("%d%d",&n,&m); for (i=2;i<=n;i++){ scanf("%d",&x); x++; Link(x,i); } for (i=1;i<=m;i++){ scanf("%d%d%d",&l,&r,&x); l++; r++; x++; q[i].num = l-1; q[i].x = -x; q[i].ans = i; q[i+m].num = r; q[i+m].x = x; q[i+m].ans = i; } sort(q+1,q+m+m+1); for (j=1;j<=m+m&&q[j].num==0;j++); for (i=1;i<=n;i++){ makeroot(1); access(i); Splay(i); tip[i]++; sum[i] += size[i]; w[i]++; while (j<=m+m && q[j].num==i){ makeroot(1); access( abs(q[j].x) ); Splay( abs(q[j].x) ); if (q[j].x>0) ans[q[j].ans] += sum[q[j].x]; else ans[q[j].ans] -= sum[-q[j].x]; j++; } } for (i=1;i<=m;i++) printf("%d\n",(ans[i]+Mod)%Mod); return 0; }