Bzoj4154--ipsc2015 Generating Synergy

Test instructions: AC Channel
Exercises
There is a very ingenious solution to this problem.
Because there are sub-tree modifications, but not the whole subtrees tree modification, directly on the DFS sequence is not possible.
Considering that it is only a child node each time it is modified, and there is a distance limit, we think that this is a problem with the depth of the point.
So we take the point of the tree to the two-dimensional plane, the horizontal axis for its DFS sequence, the ordinate for its depth, so that each time the modification corresponds to a region of the dyeing operation, this thing with Kdtree can casually engage, get a lazytag on it.

A bunch of debugging code, so it looks very long.

#include <bits/stdc++.h>//#include <conio.h>UsingnamespaceStd;typedef Long Long ll;typedef double db;constintinf=0x3f3f3f3f;intGetint () {intf=1, g=0; Char C=getchar (); while(c<' 0 '|| C>' 9 '){if(c=='-') f=-1; C=getchar ();} while(c>=' 0 '&& c<=' 9 ') g= (g<<3) + (g<<1) +c-' 0 ', C=getchar ();returnF*g;} Constintmaxn=100005; constintMod=1000000007;intDimintN,m;struct node{intd[2];int Max[2];int min[2];intL,r;intTagint Color;int& operator [] (const unsignedint& num) {returnD[num]; } BOOL operator < (const node& NO) const {returnd[dim]<no.d[dim]; } bool operator = = (const node& NO) const {returnd[0]==no.d[0] && d[1]==no.d[1]; }};node P[MAXN];intRoot;struct kdtree{node T[MAXN]; #define LC T[X].L #define RC T[X].R #definelsT[T[X].L] #define RS t[t[x].r] void update (intx) {t[x].min[0]=T[X].Max[0]=t[x][0]; T[X].min[1]=T[X].Max[1]=t[x][1];if(LC) {T[x].min[0]=min(T[x].min[0],ls.min[0]); T[X].Max[0]=Max(T[x].Max[0],ls.Max[0]); T[X].min[1]=min(T[x].min[1],ls.min[1]); T[X].Max[1]=Max(T[x].Max[1],ls.Max[1]); }if(RC) {t[x].min[0]=min(T[x].min[0],rs.min[0]); T[X].Max[0]=Max(T[x].Max[0],rs.Max[0]); T[X].min[1]=min(T[x].min[1],rs.min[1]); T[X].Max[1]=Max(T[x].Max[1],rs.Max[1]); }} void Push_down (intx) {if(!x | |!t[x].tag)return; T[X].Color=t[x].tag;ls. Tag=rs.tag=t[x].tag; t[x].tag=0; } #define MID (l+r>>1)intBuildintLintRintNow) {Dim=now; Nth_element (p+l,p+mid,p+r+1);intX=mid; T[X].Color=1; for(intI=0;i<2; i++) {t[x].min[I]=t[x].Max[I]=t[x][i]=p[x][i]; }if(L<mid) Lc=build (l,mid-1, now^1);if(R>mid) Rc=build (mid+1, r,now^1); Update (x);returnX } #undef mid node t1,t2;intCol void SetColor (intx) {if(!x)return; Push_down (x);if(T[x].min[0]>t2[0] || T[X].Max[0]<t1[0] || T[X].min[1]>t2[1] || T[X].Max[1]<t1[1])return;if(T[x].min[0]>=t1[0] && t[x].Max[0]<=t2[0] && t[x].min[1]>=t1[1] && t[x].Max[1]<=t2[1]) {t[x].tag=col;return; }if(t[x][0]>=t1[0] && t[x][0]<=t2[0] && t[x][1]>=t1[1] && t[x][1]<=t2[1]) t[x].Color=col;    SetColor (LC); SetColor (RC); } node T;intGetColorintx) {Push_down (x);if(t[0]&LT;T[X].min[0] || t[0]&GT;T[X].Max[0] || t[1]&LT;T[X].min[1] || t[1]&GT;T[X].Max[1])return 0;if(t[x]==t)returnT[X].Color;returnGetColor (LC) +getcolor (RC); } void Set (intXintYintXxintYyintCO) {t1[0]=x; t1[1]=y; t2[0]=xx; t2[1]=yy;        Col=co;    SetColor (root); }intQueryintXintY) {t[0]=x; t[1]=y;returnGetColor (root); } void Init () {memset (T,0, sizeof T); Memset (P,0, sizeof P); root=0; } Void Run (intx) {if(!x)return;        Run (LC); printf( %d,%d), (%d,%d):%d (%d) (%d,%d) \ n ", T[x].min[0],T[X].min[1],T[X].Max[0],T[X].Max[1],T[X].Color, t[x].tag,t[x][0],t[x][1]);    Run (RC); #undef LC #undef RC #undefls#undef Rs}kdtree;Vector<int> G[maxn];void Addedge (intFromintTo) {g[from].push_back (to); G[to].push_back (from);}intDEP[MAXN];intST[MAXN];intED[MAXN];intIndintFather[maxn];void DFS (intx) {st[x]=++ind; for(intI=0; I<g[x].size(); i++) {intTo=g[x][i];if(father[x]==to)Continue; dep[to]=dep[x]+1;    DFS (to); } Ed[x]=ind;}intMain () {//Freopen ("In.txt", "R", stdin);    intT=getint (); while(t--) { for(intI=0; i<maxn;i++) G[i].Clear();        Kdtree.init (); Ind=0; N=getint ();intTemp=getint (); M=getint (); for(intI=2; i<=n;i++) {father[i]=getint ();        Addedge (I,father[i]); } dep[1]=1; Dfs1);/* for (int i=1;i<=n;i++) {printf ("%d", st[i]);        } puts ("");        for (int i=1;i<=n;i++) {printf ("%d", ed[i]);        } puts ("");        for (int i=1;i<=n;i++) {printf ("%d", dep[i]);        } puts (""); */         for(intI=1; i<=n;i++) {p[i][0]=st[i]; p[i][1]=dep[i]; } root=kdtree.build (1N0);/* Kdtree.run (root);            Puts ("");        Getch (); */        intx, y;intOpt ll ans=0; for(intI=1; i<=m;i++) {x=getint ();            Y=getint (); Opt=getint ();if(!opt) {ll t=kdtree.query (st[x],dep[x]);//printf ("%d\n", t);ans+= (LL) t*(ll) I; Ans%=MoD }Else{/ * printf ("We are painting (%d,%d), (%d,%d):%d\n", st[x],dep[x],ed[x],dep[x]+y,opt);                Getch (); */Kdtree.set (st[x],dep[x],ed[x],dep[x]+y,opt); }/* Kdtree.run (root);            Puts ("");            Getch (); */} printf ("%d\n", ans); }return 0;}

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Bzoj4154--ipsc2015 Generating Synergy