Spasmodic（交互题）【树】【二分】

>Link

luogu T207969

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>Description

$n\le 500$

idea：stoorz

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>解题思路

本人写的第一道交互题QwQ

我们假定 1 为根，那就可以通过 $n-1$ 次询问第 $i$ 个点和第 $1$ 个点得到每个节点的深度，将图分层
现在我们要找到当前这层，每个点的父亲，显然这些父亲都是在上一层的
既然枚举到了这一层，那就说明前面的层的父子关系都是确定了的，我们可以处理出 $g_i$，表示上一层前 $i$ 个点与根节点的最小连通子图的大小
二分当前节点的父亲是哪个，询问上一层 $mid$ 之前所有的点，和根节点、当前节点的最小连通子图大小，如果大小 $=g_{mid}+1$，可以得出结论当前节点的父亲在 $mid$ 之前，否则在之后

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>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 510
using namespace std;

struct node
{
    
    
	int dep, id;
} a[N];
int n, fa[N], g[N], x, last, L, R, l, r, mid;
bool vis[N];

bool cmp (node aa, node bb) {
    
    return aa.dep < bb.dep;}

int main()
{
    
    
	scanf ("%d", &n);
	a[1] = (node){
    
    0, 1};
	for (int i = 2; i <= n; i++)
	{
    
    
		std::cout << 1 << " " << 2 << " " << 1 << " " << i << std::endl;
		std::cout << std::flush;
		std::cin >> x;
		a[i] = (node){
    
    x, i};
	}
	sort (a + 1, a + 1 + n, cmp);
	L = R = last = 1;
	for (int i = 2; i <= n; i++)
	{
    
    
		if (a[i].dep != a[i - 1].dep)
		{
    
    
			memset (vis, 0, sizeof (vis));
			memset (g, 0, sizeof (g));
			for (int j = last; j <= i - 1; j++)
			{
    
    
				g[j] = g[j - 1];
				for (int now = a[j].id; now > 1; now = fa[now])
			      if (!vis[now]) vis[now] = 1, g[j]++;
				  else break;
			}
			L = last, R = i - 1, last = i;
		}
		l = L, r = R;
		while (l < r)
		{
    
    
			mid = (l + r) / 2;
			std::cout << 1 << " " << mid - L + 3;
			std::cout << " " << 1 << " " << a[i].id;
			for (int j = L; j <= mid; j++)
			  std::cout << " " << a[j].id;
			std::cout << std::endl;
			std::cout << std::flush;
			std::cin >> x;
			if (x == g[mid] + 1) r = mid;
			else l = mid + 1;
		}
		fa[a[i].id] = a[r].id;
	}
	std::cout << 2;
	for (int i = 2; i <= n; i++)
	  std::cout << " " << fa[i] << " " << i;
	std::cout << std::endl;
	std::cout << std::flush;
	return 0;
}