传送门
文章目录
- 题意:
- 思路:
题意:
给你一颗树,每次可以询问两个点的lcalcalca,询问次数不能超过⌊n2⌋\left \lfloor \frac{n}{2} \right \rfloor⌊2n⌋,求它的根。
n≤1e3n\le1e3n≤1e3。
思路:
看到询问次数≤⌊n2⌋\le \left \lfloor \frac{n}{2} \right \rfloor≤⌊2n⌋,比较容易想到每次询问两个点,那么询问哪两个点呢?比较容易想到询问叶子节点,所以我们只需要每次询问两个叶子的lcalcalca,如果他们的lcalcalca等于其中任意一个叶子,那说明根节点就是这个点了,否则就删掉这两个点,将新的叶子节点加入。一直到最后只剩一个点的时候,这个点就是lcalcalca了。
// Problem: D. Kuroni and the Celebration
// Contest: Codeforces - Ozon Tech Challenge 2020 (Div.1 + Div.2, Rated, T-shirts + prizes!)
// URL: https://codeforces.com/contest/1305/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<assert.h>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int d[N],cnt;
vector<int>v[N],temp;
int st[N];int query(int a,int b) {printf("? %d %d\n",a,b); fflush(stdout);int x; scanf("%d",&x);return x;
}int solve() {queue<int>q;for(int i=1;i<=n;i++) if(d[i]==1) q.push(i),st[i]=1;while(q.size()>1) {int a=q.front(); q.pop();int b=q.front(); q.pop();int root=query(a,b);if(root==a||root==b) return root;for(auto x:v[a]) if(!st[x]&&(--d[x]==1)) q.push(x);for(auto x:v[b]) if(!st[x]&&(--d[x]==1)) q.push(x);st[a]=st[b]=1; }for(int i=1;i<=n;i++) if(!st[i]) return i;return -1;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);scanf("%d",&n);for(int i=1;i<=n-1;i++) {int a,b; scanf("%d%d",&a,&b);d[a]++; d[b]++;v[a].pb(b); v[b].pb(a);}printf("! %d\n",solve()); fflush(stdout);return 0;
}
/**/