Codeforces Round 865 (Div. 2) D. Sum GraphD - Sum Graph 题目内容

题目内容

This is an interactive problem. There is a hidden permutation $p_1, p_2, \dots, p_n$ . Consider an undirected graph with $n$ nodes only with no edges. You can make two types of queries: Specify an integer $x$ satisfying $2 \le x \le 2n$ . For all integers $i$ ( $1 \le i \le n$ ) such that $1 \le x-i \le n$ , an edge between node $i$ and node $x-i$ will be added. Query the number of edges in the shortest path between node $p_i$ and node $p_j$ . As the answer to this question you will get the number of edges in the shortest path if such a path exists, or $-1$ if there is no such path. Note that you can make both types of queries in any order. Within $2n$ queries (including type $1$ and type $2$ ), guess two possible permutations, at least one of which is $p_1, p_2, \dots, p_n$ . You get accepted if at least one of the permutations is correct. You are allowed to guess the same permutation twice. A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ( $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ( $n=3$ but there is $4$ in the array).

中文大意

给我们n个点没有边的无向图这时有操作一和操作二

输入一个整数 x 将 i 和 x - i连边并且 i和x - i都在[1, n] 的范围内

输入两个整数i, j若存在路径返回i->j的最短路否则返回-1
让我们在2n次操作内去猜测可能的序列

解法

当我们进行两次x 为n,n+1的操作一我们会发现这个图被连成了一条直线并且端点是n和(n + 1) / 2所有我们可以通过n - 1次查询去求得端点然后再通过n-1次查询去确定位置，最后输出即可

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
// #define endl '\n'
#define pb push_back
#define NO cout << "NO" << endl;
#define YES cout << "YES" << endl;
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<VI> VII;
ll MOD;
ll powmod(ll a,ll b) {ll res=1;a%=MOD; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;}
mt19937 mrand(random_device{}()); 
const int N = 2e5 + 10;
/*
    3 2 6 1 4 5 2次
    1 2 
*/
void solve()
{
    int n; cin >> n;
    VII e(n + 1);
    int k = n;
    for (int i = 1; i <= n; i++) {
        if (i == k) {
            k--;
            continue;
        }
        e[i].pb(k--);
    }
    k = n - 1;
    for (int i = 1; i < n; i++) {
        if (i == k) {
            k--;
            continue;
        }
        e[i].pb(k--);
    }
    VI a(n + 1);
    function<void(int, int, int)> dfs = [&] (int u, int fa, int num) {
        a[num] = u;
        // cout << num << endl;
        for (auto v : e[u]) if (v != fa) {
            dfs(v, u, num + 1);
        }
    };
    cout << "+ " << n + 1 << endl;
    int x; cin >> x;
    if (x == -2) return;
    cout << "+ " << n << endl;
    cin >> x;
    if (x == -2) return;
    dfs(n, 0, 1);
    // rep (i, 1, n) cout << a[i] << " \n"[i==n];
    int mx = 0, d = 0;
    for (int i = 2; i <= n; i++) {
        cout << "? 1" << ' ' << i << endl;
        cin >> x;
        if (x == -2) return;
        if (x > mx) {
            mx = x;
            d = i;
        }
    }
    VI res(n + 1), ans(n + 1);
    rep (i, 1, n) {
        if (i == d) continue;
        cout << "? " << d << ' ' << i << endl;
        cin >> x;
        if (x == -2) return;
        res[i] = a[x + 1];
        // cout << i << " res=" << a[x + 1] << endl; 
        ans[i] = a[n - x];
        // cout << i << " ans=" << a[n - x] << endl;
    }
    res[d] = a[1], ans[d] = a[n];
    cout << "! ";
    rep (i, 1, n) cout << res[i] << ' ';
    rep (i, 1, n) cout << ans[i] << ' ';
    cout << endl;
    cin >> x;
    if (x == -2) return;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;cin >> T;
    while ( T -- )
    solve();
    return 0;
}