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You are given a tree (an acyclic undirected connected graph) with n nodes. The tree nodes are numbered from 1 to n.
Each node has a color, white or black. All the nodes are black initially.
We will ask you to perfrom some instructions of the following form:
The first line contains a number n denoted how many nodes in the tree(1 ≤ n ≤ 105). The next n - 1 lines, each line has two numbers (u, v) describe a edge of the tree(1 ≤ u, v ≤ n). The next line contains a number m denoted how many operations we are going to process(1 ≤ m ≤ 105). The next m lines, each line describe a operation (t, u) as we mentioned above(0 ≤ t ≤ 1, 1 ≤ u ≤ n).
For each query operation, output the corresponding result.
这道题常数卡的有点紧，我树链剖分用一棵线段树存就TLE了，每个链分别建线段树才行。
考虑将每一个同色块的答案保存在这一块深度最浅的那一个点（这是一个很好的思路），我们考虑如何维护即可，对于每一个点，我们维护f[now][0/1]表示当前点如果取白色/黑色，所在的子树中与这个点同色的联通块大小。
每次颜色修改只会影响到当前点到根节点路径上的一段。而且还是路径加减一个数，这可以用链剖维护。
询问时只用跳到当前联通块最上方的点，然后输出该点所存的f值即可。
AC600了，lalala～～
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define MAXN 101000 #define MAXV MAXN*2 #define MAXE MAXV*2 #define MAXT MAXN*4 #define lch sgt[now].lc #define rch sgt[now].rc #define smid ((l+r)>>1) #define INF 0x3f3f3f3f struct Edge { int np; Edge *next; }E[MAXE],*V[MAXV]; int tope=-1; void addedge(int x,int y) { E[++tope].np=y; E[tope].next=V[x]; V[x]=&E[tope]; } int q[MAXN]; int pnt[MAXN]; int siz[MAXN]; int son[MAXN],top[MAXN],pos[MAXN],apos[MAXN],dfstime; void bfs(int now) { int head=-1,tail=0; Edge *ne; q[0]=now; pnt[now]=0; while (head<tail) { now=q[++head]; for (ne=V[now];ne;ne=ne->next) { if (ne->np==pnt[now])continue; pnt[ne->np]=now; q[++tail]=ne->np; } } for (int i=tail;i>=0;i--) { now=q[i]; siz[now]=1; int mxsiz=0; for (ne=V[now];ne;ne=ne->next) { if (ne->np==pnt[now])continue; siz[now]+=siz[ne->np]; if (siz[ne->np]>mxsiz) { mxsiz=now; son[now]=ne->np; } } } } int stack[MAXN],tops=-1; void dfs(int now) { Edge *ne; stack[++tops]=now; top[now]=now; while (~tops) { now=stack[tops--]; pos[now]=++dfstime; apos[dfstime]=now; for (ne=V[now];ne;ne=ne->next) { if (ne->np==pnt[now] || ne->np==son[now])continue; stack[++tops]=ne->np; top[ne->np]=ne->np; } if (son[now]) { stack[++tops]=son[now]; top[son[now]]=top[now]; } } } int col[MAXN]; int ptra[MAXN]; struct sgt_node { int lc,rc; int sum[2]; int pls[2]; int val[2]; }sgt[MAXT]; int topt=0; void make_plus(int now,int c,int d) { sgt[now].val[c]+=d; sgt[now].pls[c]+=d; } void down(int now) { if (sgt[now].pls[0]) { make_plus(lch,0,sgt[now].pls[0]); make_plus(rch,0,sgt[now].pls[0]); sgt[now].pls[0]=0; } if (sgt[now].pls[1]) { make_plus(lch,1,sgt[now].pls[1]); make_plus(rch,1,sgt[now].pls[1]); sgt[now].pls[1]=0; } } int Build_sgt(int l,int r) { int now=++topt; sgt[now].val[0]=sgt[now].val[1]=sgt[now].pls[0]=sgt[now].pls[1]=0; sgt[now].sum[1]=1; sgt[now].sum[0]=0; if (l==r) { ptra[l]=now; sgt[now].val[1]=siz[apos[l]]; sgt[now].val[0]=1; return now; } lch=Build_sgt(l,smid); rch=Build_sgt(smid+1,r); return now; } pair<int,int> Query_sgt(int now,int l,int r,int pos) { if (l==r) return make_pair(sgt[now].val[0],sgt[now].val[1]); down(now); if (pos<=smid) return Query_sgt(lch,l,smid,pos); else return Query_sgt(rch,smid+1,r,pos); } void Modify_sgt(int now,int l,int r,int x,int y,int c,int d) { if (l==x && r==y) { make_plus(now,c,d); return ; } down(now); if (y<=smid) return Modify_sgt(lch,l,smid,x,y,c,d); else if (smid<x) return Modify_sgt(rch,smid+1,r,x,y,c,d); else { Modify_sgt(rch,smid+1,r,smid+1,y,c,d); Modify_sgt(lch,l,smid,x,smid,c,d); } } void Modify_sgt2(int now,int l,int r,int pos) { if (l==r) return swap(sgt[now].sum[0],sgt[now].sum[1]); down(now); if (pos<=smid) Modify_sgt2(lch,l,smid,pos); else Modify_sgt2(rch,smid+1,r,pos); sgt[now].sum[0]=sgt[lch].sum[0]+sgt[rch].sum[0]; sgt[now].sum[1]=sgt[lch].sum[1]+sgt[rch].sum[1]; } int Scan_sgt(int now,int l,int r,int x,int y,int c) { if (l==x && r==y) { if (sgt[now].sum[c]==(r-l+1)) { return l; }else if (sgt[now].sum[c]==0) { return -1; }else { down(now); int ret=Scan_sgt(rch,smid+1,r,smid+1,y,c); if (ret==smid+1) { ret=Scan_sgt(lch,l,smid,x,smid,c); if (ret==-1)return smid+1; else return ret; }else return ret; } } down(now); if (y<=smid) return Scan_sgt(lch,l,smid,x,y,c); else if (smid<x) return Scan_sgt(rch,smid+1,r,x,y,c); else { int ret=Scan_sgt(rch,smid+1,r,smid+1,y,c); if (ret==smid+1) { ret=Scan_sgt(lch,l,smid,x,smid,c); if (ret==-1)return smid+1; else return ret; }else return ret; } } int spos[MAXN],tpos[MAXN]; int troot[MAXN]; int Swim_up(int x) { int rpos=pos[x]; int c=col[x]; while (x) { int y=Scan_sgt(troot[top[x]],spos[top[x]],tpos[top[x]],pos[top[x]],pos[x],c); if (y==-1)break; else if (y!=pos[top[x]]) { rpos=y;break; }else { rpos=y; x=pnt[top[x]]; } } return apos[rpos]; } int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); int n,m; int x,y,z; scanf("%d",&n); for (int i=1;i<n;i++) { scanf("%d%d",&x,&y); addedge(x,y); addedge(y,x); } bfs(1); dfs(1); for (int i=1;i<=n;i++) spos[i]=INF,tpos[i]=-INF; for (int i=1;i<=n;i++) spos[top[i]]=min(spos[top[i]],pos[i]); for (int i=1;i<=n;i++) tpos[top[i]]=max(tpos[top[i]],pos[i]); for (int i=1;i<=n;i++) if (top[i]==i) troot[i]=Build_sgt(spos[i],tpos[i]); scanf("%d",&m); int opt; for (int i=1;i<=n;i++)col[i]=1; for (int i=0;i<m;i++) { scanf("%d%d",&opt,&x); if (opt==0) { int rpt=Swim_up(x); pair<int,int> res=Query_sgt(troot[top[rpt]],spos[top[rpt]],tpos[top[rpt]],pos[rpt]); if (col[x]==0) printf("%d\n",res.first); else printf("%d\n",res.second); }else { pair<int,int> res=Query_sgt(troot[top[x]],spos[top[x]],tpos[top[x]],pos[x]); int p=pnt[x]; int c=col[p]; int d,d2; if (col[p]==1 && col[x]==1)d=-res.second,d2=res.first; else if (col[p]==1 && col[x]==0)d=res.second,d2=-res.first; else if (col[p]==0 && col[x]==1)d=res.first,d2=-res.second; else d=-res.first,d2=res.second; if (p) { sgt[ptra[pos[p]]].val[c^1]+=d2; int a=Swim_up(p); a=pnt[a]; if (!a)a=1; while (true) { if (top[p]==top[a]) { Modify_sgt(troot[top[a]],spos[top[a]],tpos[top[a]],pos[a],pos[p],c,d); break; } Modify_sgt(troot[top[p]],spos[top[p]],tpos[top[p]],pos[top[p]],pos[p],c,d); p=pnt[top[p]]; } } Modify_sgt2(troot[top[x]],spos[top[x]],tpos[top[x]],pos[x]); col[x]^=1; } } }
bzoj 3637: Query on a tree VI 树链剖分 && AC600
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原文地址：http://www.cnblogs.com/mhy12345/p/4748393.html
