这题有点意思
转化问题，我们把询问区间的点到根链加，再查询询问点到根的权值和就是每个询问的答案。
然后如果你数据结构没学傻只需要差分一下就可以扫一遍出解了
1 #include<cstdio> 2 #include<vector> 3 #include<cstring> 4 #include<algorithm> 5 #define vpii vector<pair<int,int> > 6 #define vpit vector<pair<int,int> >::iterator 7 using namespace std; 8 const int N=50005,M=200005,mod=201314; 9 int p[N],noww[M],goal[M],val[M],laz[M],siz[N]; 10 int far[N],dep[N],imp[N],dfn[N],top[N],ans[N]; 11 int n,q,t1,t2,t3,cnt,tot; vpii lp[N],rp[N]; 12 void Add(int &x,int y) 13 { 14 x+=y; 15 if(x>=mod) x-=mod; 16 } 17 void Link(int f,int t) 18 {//printf("Link %d with %d",f,t); 19 noww[++cnt]=p[f]; 20 goal[cnt]=t,p[f]=cnt; 21 noww[++cnt]=p[t]; 22 goal[cnt]=f,p[t]=cnt; 23 } 24 void Release(int nde,int l,int r) 25 { 26 if(laz[nde]) 27 { 28 int mid=(l+r)/2,ls=2*nde,rs=2*nde+1,&lz=laz[nde]; 29 val[ls]+=lz*(mid-l+1),laz[ls]+=lz; 30 val[rs]+=lz*(r-mid),laz[rs]+=lz,lz=0; 31 } 32 } 33 void Change(int nde,int l,int r,int ll,int rr,int tsk) 34 { 35 if(l>rr||r<ll) 36 return ; 37 else if(l>=ll&&r<=rr) 38 val[nde]+=tsk*(r-l+1),laz[nde]+=tsk; 39 else 40 { 41 int mid=(l+r)/2,ls=2*nde,rs=2*nde+1; Release(nde,l,r); 42 Change(ls,l,mid,ll,rr,tsk),Change(rs,mid+1,r,ll,rr,tsk); 43 val[nde]=val[ls]+val[rs]; 44 } 45 } 46 int Query(int nde,int l,int r,int ll,int rr) 47 { 48 if(l>rr||r<ll) 49 return 0; 50 else if(l>=ll&&r<=rr) 51 return val[nde]; 52 else 53 { 54 int mid=(l+r)/2,ls=2*nde,rs=2*nde+1; Release(nde,l,r); 55 return Query(ls,l,mid,ll,rr)+Query(rs,mid+1,r,ll,rr); 56 } 57 } 58 void DFS(int nde,int fth,int dth) 59 { 60 int tmp=0; 61 siz[nde]=1,far[nde]=fth,dep[nde]=dth; 62 for(int i=p[nde];i;i=noww[i]) 63 if(goal[i]!=fth) 64 { 65 DFS(goal[i],nde,dth+1); 66 siz[nde]+=siz[goal[i]]; 67 if(siz[goal[i]]>tmp) 68 tmp=siz[goal[i]],imp[nde]=goal[i]; 69 } 70 } 71 void Decompose(int nde,int tpp) 72 { 73 dfn[nde]=++tot,top[nde]=tpp; 74 if(imp[nde]) 75 { 76 Decompose(imp[nde],tpp); 77 for(int i=p[nde];i;i=noww[i]) 78 if(goal[i]!=imp[nde]&&goal[i]!=far[nde]) 79 Decompose(goal[i],goal[i]); 80 } 81 } 82 void Chain_Change(int x,int y,int v) 83 { 84 while(top[x]!=top[y]) 85 { 86 if(dep[top[x]]<dep[top[y]]) swap(x,y); 87 Change(1,1,n,dfn[top[x]],dfn[x],v),x=far[top[x]]; 88 } 89 if(dep[x]>dep[y]) swap(x,y); 90 Change(1,1,n,dfn[x],dfn[y],v); 91 } 92 int Chain_Query(int x,int y) 93 { 94 int ret=0; 95 while(top[x]!=top[y]) 96 { 97 if(dep[top[x]]<dep[top[y]]) swap(x,y); 98 ret+=Query(1,1,n,dfn[top[x]],dfn[x]),x=far[top[x]]; 99 } 100 if(dep[x]>dep[y]) swap(x,y); 101 ret+=Query(1,1,n,dfn[x],dfn[y]); return ret; 102 } 103 int main() 104 { 105 scanf("%d%d",&n,&q); 106 for(int i=2;i<=n;i++) 107 scanf("%d",&t1),Link(t1+1,i); 108 for(int i=1;i<=q;i++) 109 { 110 scanf("%d%d%d",&t1,&t2,&t3); 111 lp[t1].push_back(make_pair(i,t3+1)); 112 rp[t2+1].push_back(make_pair(i,t3+1)); 113 } 114 DFS(1,0,1),Decompose(1,1); 115 for(int i=1;i<=n;i++) 116 { 117 Chain_Change(1,i,1); 118 for(vpit it=lp[i].begin();it!=lp[i].end();it++) 119 ans[it->first]-=Chain_Query(1,it->second);//,printf("%d-=%d ",it->first,Chain_Query(1,it->second)); 120 for(vpit it=rp[i].begin();it!=rp[i].end();it++) 121 ans[it->first]+=Chain_Query(1,it->second);//,printf("%d+=%d ",it->first,Chain_Query(1,it->second)); 122 } 123 for(int i=1;i<=q;i++) 124 printf("%d ",ans[i]%mod); 125 return 0; 126 }