Topic description
Given a tree with N points (N-1 edges), the nodes are white and black, and the initial is all white
There are two operations:
0 i : Change the color of a point (the original black becomes white, the original is White becomes black)
1 v : ask for the first black point on the path from 1 to v, if not, output -1
Input format:
The first line N, Q, representing N points and Q operations,
the second line to the Nth line N-1 undirected edges
and then Q lines, each line is an operation "0 i" or "1 v" (1 ≤ i, v ≤ N).
Output format:
output result for each 1 v operation
input sample
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
Sample output
-1
8
-1
2
-1
illustrate
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.
***************************
Topic Analysis:
Use the point weight of 0 to represent white, and the point weight of 1 to represent black
. After the tree is cut,
find the point weight closest to node 1 and the chain from u to top[u] that is >=1.
Let ll=num[ top[ u] ], rr=num[u]
Then in this interval, binary search for the point with the smallest num value and the weight of the point with the weight of 1 output
****************
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
int read()
{
int f=1,x=0;
char ss=getchar();
while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
return f*x;
}
int n,m;
struct node{int v,nxt;}E[300010];
int head[200010],tot;
int dep[200010],fa[200010],son[200010],size[200010];
int top[200010],num[500010],pos[500010],cnt;
int sum[500010];
void add(int u,int v)
{
E[++tot].nxt=head[u];
E[tot].v=v;
head[u]=tot;
}
void dfs1(int u,int pa)
{
size[u]=1;
for(int i=head[u];i;i=E[i].nxt)
{
int v=E[i].v;
if(v==pa) continue;
dep[v]=dep[u]+1; fa[v]=u;
dfs1(v,u);
size[u]+=size[v];
if(size[v]>size[son[u]]) son[u]=v;
}
}
void dfs2(int u,int tp)
{
top[u]=tp; num[u]=++cnt; pos[cnt]=u;
if(son[u]) dfs2(son[u],tp);
for(int i=head[u];i;i=E[i].nxt)
{
int v=E[i].v;
if(v==fa[u]||v==son[u]) continue;
dfs2(v,v);
}
}
void update(int u,int s,int t,int p)
{
if(s==t){sum[p]^=1; return;}//修改颜色直接取反
int mid=s+t>>1;
if(u<=mid) update(u,s,mid,p<<1);
else update(u,mid+1,t,p<<1|1);
sum[p]=sum[p<<1]+sum[p<<1|1];
}
int get(int ll,int rr,int s,int t,int p)
{
if(ll<=s&&t<=rr) return sum[p];
int mid=s+t>>1;
int ans=0;
if(ll<=mid) ans+=get(ll,rr,s,mid,p<<1);
if(rr>mid) ans+=get(ll,rr,mid+1,t,p<<1|1);
return ans;
}
int check(int ll,int rr)
{
if(ll==rr) return ll;
int mid=ll+rr>>1;
int tp=get(ll,mid,1,n,1);
if(tp>=1) return check(ll,mid);//尽量靠近1号点
else return check(mid+1,rr);
}
int query(int u)
{
int ll=0,rr=0;
while(top[u]!=1)
{
int tp=get(num[top[u]],num[u],1,n,1);
if(tp>=1) ll=num[top[u]],rr=num[u];//找到一个权值和>=1的区间先记录
u=fa[top[u]];
}
int tp=get(1,num[u],1,n,1);//以1号点为链顶的最后一条链
if(tp>=1) ll=1,rr=num[u];
if(ll==0&&rr==0) return -1;//没有黑点
else return pos[check(ll,rr)];//二分查找
}
int main()
{
n=read();m=read();
for(int i=1;i<n;++i)
{
int u=read(),v=read();
add(u,v);add(v,u);
}
dep[1]=1;
dfs1(1,0);dfs2(1,1);
while(m--)
{
int k=read(),u=read();
if(k==0) update(num[u],1,n,1);
else if(k==1) printf("%d\n",query(u));
}
return 0;
}