bzoj4448 [Scoi2015] Intelligence transmission

The first question does not explain, for the second question processing, may use the CDQ divide the treatment, assumes the division rule the question interval is [l,r], then we to the marking in [L,mid] The modification operation assigns a weight value, because in the current [L,r] [L,mid] 's modification operation only to [MID+1,R] the inquiry operation , so I modify the operation at least m-i time, so the weighted value is m-i, and for the [mid+1,r] interval of the query operation, also gives a weight w-m, where w is the value of the inquiry, we can preprocess the tree's DFS sequence and maintain a tree array, This allows us to insert or inquire about these operations from the weights to the small, Time complexity O (NLOGNLOGN).

Code, running time is almost the bottom line. (looking at the other blogger's puzzle seems to me a complex ... )

1#include <cstdio>2#include <algorithm>3 #defineN 5000104 #defineLB (x) (x&-x)5 using namespacestd;6 intDp,p[n],pre[n],tt[n],n,a,i,m,o,c[n],deep[n],fa[n];7 intL[n],r[n],ans[n];8 ints[n][ -];9 structg{Ten     inttyp,l,r,w; One }b[n]; A structgg{ -     intid,v; - }w[n]; the voidccintXintW) - { -      while(x<=o) -     { +c[x]+=W; -x+=lb (x); +     } A } at voidDfsintx) - { -     intI=P[x]; -l[x]=++o; -      while(i) -     { infa[tt[i]]=x; -deep[tt[i]]=deep[x]+1; to DFS (Tt[i]); +I=Pre[i]; -     } ther[x]=++o; * } $ intLcaintXinty)Panax Notoginseng { -     if(Deep[x]>deep[y]) x^=y^=x^=y; the     inti; +      for(i= +; i>=0; i--) A     { the         if(Deep[y]-deep[x]>= (1<<i)) +         { -y=S[y][i]; $         } $     } -     if(x==y)returnx; -      for(i= +; i>=0; i--) the     { -         if(s[x][i]!=S[y][i])Wuyi         { thex=S[x][i]; -y=S[y][i]; Wu         } -     } About     returnFa[x]; $ } - intSumintx) - { -     intans=0; A      while(x) +     { theans+=C[x]; -x-=lb (x); $     } the     returnans; the } the voidLinkintXinty) the { -dp++;p re[dp]=p[x];p [x]=dp;tt[dp]=y; in } the BOOLCMP (GG a,gg b) the { About     if(a.v==B.V) the     returnA.id>b.id; the     returnA.v>B.V; the } + voidSolveintLintR) - { the     intm,i,tot=0;Bayi     if(L&GT;=R)return; theM= (l+r) >>1; the      for(i=l;i<=m;i++) -     if(b[i].typ==2) -     { theW[++tot].id=i; thew[tot].v=m-i; the     } the      for(i=m+1; i<=r;i++) -     if(b[i].typ==1) the     { theW[++tot].id=i; thew[tot].v=b[i].w-(I-m);94     } theSort (w+1, w+1+tot,cmp); the      the      for(i=1; i<=tot;i++)98     if(w[i].id<=m) About     { -CC (L[B[W[I].ID].W],1);101CC (r[b[w[i].id].w],-1);102     }103     Else104     { the         intu=B[W[I].ID].L;106         intv=B[W[I].ID].R;107         intLca=LCA (U,V);108Ans[w[i].id]+=sum (L[u]) +sum (L[v])-sum (L[lca])-sum (l[lca]-1);109     } the      for(i=1; i<=tot;i++)111     if(w[i].id<=m) the     {113CC (l[b[w[i].id].w],-1); theCC (R[B[W[I].ID].W],1); the     } the 117Solve (l,m); Solve (m+1, R);118 }119  - intMain ()121 {122scanf"%d",&n);123      for(i=1; i<=n;i++)124     { thescanf"%d",&a);126         if(a) link (a,i);127     } -scanf"%d",&m);129      for(i=1; i<=m;i++) the     {131scanf"%d",&B[i].typ); the         if(b[i].typ==1)133scanf"%d%d%d",&b[i].l,&b[i].r,&B[I].W);134         Elsescanf"%d",&B[I].W);135     }136Dfs1);137      for(i=1; i<=n;i++) s[i][0]=Fa[i];138      for(intH=1;h< -; h++)139     { $          for(i=1; i<=n;i++)141         {142s[i][h]=s[s[i][h-1]][h-1];143         }144     }145Solve1, m);146      for(i=1; i<=m;i++)147     if(b[i].typ==1)148     {149         intans=LCA (B[I].L,B[I].R); Maxans=deep[b[i].l]+deep[b[i].r]-2*deep[ans]+1;151printf"%d%d\n", Ans,ans[i]); the     }153}

bzoj4448 [Scoi2015] Intelligence transmission