SPOJ QTREE3 - Query on a tree again!
You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N. In the start, the color of any node in the tree is white.
We will ask you to perfrom some instructions of the following form:
- 0 i : change the color of the i-th node (from white to black, or from black to white);
or - 1 v : ask for the id of the first black node on the path from node 1 to node v. if it doesn't exist, you may return -1 as its result.
Input
In the first line there are two integers N and Q.
In the next N-1 lines describe the edges in the tree: a line with two integers a b denotes an edge between a and b.
The next Q lines contain instructions "0 i" or "1 v" (1 ≤ i, v ≤ N).
Output
For each "1 v" operation, write one integer representing its result.
Example
Input: 9 8 1 2 1 3 2 4 2 9 5 9 7 9 8 9 6 8 1 3 0 8 1 6 1 7 0 2 1 9 0 2 1 9 Output: -1 8 -1 2 -1
Constraints & Limits
There are 12 real input files.
For 1/3 of the test cases, N=5000, Q=400000.
For 1/3 of the test cases, N=10000, Q=300000.
For 1/3 of the test cases, N=100000, Q=100000.
一棵树，点数<=100000，初始全是白点，支持两种操作：
1.将某个点的颜色反色。
2.询问某个点至根节点路径上第一个黑点是哪个。
树链剖分
注意到询问的链都是(1,v)
在线段树上维护区间内深度最浅的黑色点位置即可，注意线段树结点到原树的反向映射
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<queue> 6 using namespace std; 7 const int INF=1e9; 8 const int mxn=100010; 9 int read(){ 10 int x=0,f=1;char ch=getchar(); 11 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();} 13 return x*f; 14 } 15 struct edge{int v,nxt;}e[mxn<<1]; 16 int hd[mxn],mct=0; 17 int n,q; 18 void add_edge(int u,int v){ 19 e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;return; 20 } 21 struct node{ 22 int fa,son; 23 int top,size; 24 int w; 25 }t[mxn]; 26 int sz=0; 27 int dep[mxn]; 28 int id[mxn]; 29 void DFS1(int u,int fa){ 30 dep[u]=dep[fa]+1; 31 t[u].size=1; 32 for(int i=hd[u];i;i=e[i].nxt){ 33 int v=e[i].v;if(v==fa)continue; 34 t[v].fa=u; 35 DFS1(v,u); 36 t[u].size+=t[v].size; 37 if(t[v].size>t[t[u].son].size) 38 t[u].son=v; 39 } 40 return; 41 } 42 void DFS2(int u,int top){ 43 t[u].w=++sz;t[u].top=top; 44 id[sz]=u; 45 if(t[u].son)DFS2(t[u].son,top); 46 for(int i=hd[u];i;i=e[i].nxt){ 47 int v=e[i].v; 48 if(v==t[u].fa || v==t[u].son)continue; 49 DFS2(v,v); 50 } 51 return; 52 } 53 struct SGT{ 54 int ps; 55 int c; 56 }st[mxn<<2]; 57 void Build(int l,int r,int rt){ 58 st[rt].ps=INF; 59 if(l==r)return; 60 int mid=(l+r)>>1; 61 Build(l,mid,rt<<1);Build(mid+1,r,rt<<1|1); 62 return; 63 } 64 void update(int p,int l,int r,int rt){ 65 if(l==r){ 66 st[rt].c^=1; 67 st[rt].ps=(st[rt].c)?l:INF; 68 return; 69 } 70 int mid=(l+r)>>1; 71 if(p<=mid)update(p,l,mid,rt<<1); 72 else update(p,mid+1,r,rt<<1|1); 73 st[rt].ps=min(st[rt<<1].ps,st[rt<<1|1].ps); 74 return; 75 } 76 int query(int L,int R,int l,int r,int rt){ 77 if(L<=l && r<=R){return st[rt].ps;} 78 int mid=(l+r)>>1; 79 int res=INF; 80 if(L<=mid)res=min(res,query(L,R,l,mid,rt<<1)); 81 if(R>mid && res>mid)res=min(res,query(L,R,mid+1,r,rt<<1|1)); 82 return res; 83 } 84 int Qt(int x,int y){ 85 int res=INF; 86 while(t[x].top!=t[y].top){ 87 if(dep[t[x].top]<dep[t[y].top])swap(x,y); 88 res=min(res,query(t[t[x].top].w,t[x].w,1,n,1)); 89 x=t[t[x].top].fa; 90 } 91 if(dep[x]>dep[y])swap(x,y); 92 res=min(res,query(t[x].w,t[y].w,1,n,1)); 93 return res; 94 } 95 int main(){ 96 int i,j,u,v; 97 n=read();q=read(); 98 for(i=1;i<n;i++){ 99 u=read();v=read(); 100 add_edge(u,v); 101 add_edge(v,u); 102 } 103 DFS1(1,0); 104 DFS2(1,1); 105 Build(1,n,1); 106 while(q--){ 107 u=read();v=read(); 108 if(!u) 109 update(t[v].w,1,n,1); 110 else{ 111 int ans=Qt(1,v); 112 if(ans>=INF){printf("-1\n");continue;} 113 ans=id[ans]; 114 printf("%d\n",ans); 115 } 116 } 117 return 0; 118 }
