CF1174F Ehab and the Big Finale(交互+剖分)

做法

\(x\)为隐藏节点,\(dep_x=d(1,x)\)
\((1)\)\(u=1\)

\((2)\):重链剖分,比如\(v\)\(u\)的重链底部,查询\(dis(x,v)\)的长度,\(y=lca(v,x)\)且在重链上,\(dis(x,v)=dep_v+dep_x-2*dep_y,dep_y=(dep_v+dep_x-dis(x,v))/2\),则我们可以找到\(y\)

\((3)\):但\(dep_y=dep_x\)时,\(y\)为答案,退出

\((4)\):找到\(y\)后,查询\(sec=(y,x)\)上的第二个节点,\(u=sec\)返回\((2)\)

code

#include<bits/stdc++.h>
typedef int LL;
const LL maxn=1e6+9;
inline LL Read(){
    LL x(0),f(1); char c=getchar();
    while(c<'0' || c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0' && c<='9'){
        x=(x<<3)+(x<<1)+c-'0'; c=getchar();
    }return x*f;
}
struct node{
	LL to,nxt;
}dis[maxn];
LL n,num;
LL head[maxn],size[maxn],tail[maxn],dep[maxn],fa[maxn],son[maxn];
inline void Add(LL u,LL v){
	dis[++num]=(node){v,head[u]}; head[u]=num;
}
void Dfs1(LL u){
	size[u]=1;
	for(LL i=head[u];i;i=dis[i].nxt){
		LL v(dis[i].to);
		if(v==fa[u]) continue;
		fa[v]=u; dep[v]=dep[u]+1;
		Dfs1(v); size[u]+=size[v];
		if(size[son[u]]<size[v]) son[u]=v;
	}
}
void Dfs2(LL u,LL f){
	if(son[u]) Dfs2(son[u],f);
	for(LL i=head[u];i;i=dis[i].nxt){
		LL v(dis[i].to);
		if(v==fa[u] || v==son[u]) continue;
		Dfs2(v,v);
	}
	if(!tail[f]) tail[f]=u;
}
LL Query(LL u,LL len){
	if(!len) return u;
	return Query(fa[u],len-1);
}
inline LL Query1(LL x){
	printf("d %d\n",x);
	fflush(stdout);
	LL ret; ret=Read();
	return ret;
}
inline LL Query2(LL x){
	printf("s %d\n",x);
	fflush(stdout);
	LL ret; ret=Read();
	return ret;
}
int main(){
	n=Read();
	for(LL i=1;i<n;++i){
		LL u(Read()),v(Read());
		Add(u,v); Add(v,u);
	}
	Dfs1(1); Dfs2(1,1);
	LL depx(Query1(1));
	LL u(1),v(tail[u]);
	while(true){
		LL dis_vx(Query1(v));
		LL sum(dep[v]+depx-dis_vx);
		LL depy(sum>>1);
		LL len(dep[v]-depy);
		LL y(Query(v,len));
		if(depx==depy){
			printf("! %d\n",y);
			fflush(stdout);
			return 0;
		}
		u=Query2(y); v=tail[u];
	}
}
posted @ 2019-06-04 14:54  y2823774827y  阅读(208)  评论(0编辑  收藏  举报