Codeforces Round #651 (Div. 2) F2. The Hidden Pair (Hard Version) (二分+剪枝)

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2020/06/24 20:03
阅读数 79

F2. The Hidden Pair (Hard Version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Note that the only difference between the easy and hard version is the constraint on the number of queries. You can make hacks only if all versions of the problem are solved.

This is an interactive problem.

You are given a tree consisting of nn nodes numbered with integers from 11 to nn. Ayush and Ashish chose two secret distinct nodes in the tree. You need to find out both the nodes. You can make the following query:

  • Provide a list of nodes and you will receive a node from that list whose sum of distances to both the hidden nodes is minimal (if there are multiple such nodes in the list, you will receive any one of them). You will also get the sum of distances of that node to the hidden nodes.

Recall that a tree is a connected graph without cycles. The distance between two nodes is defined as the number of edges in the simple path between them.

More formally, let's define two hidden nodes as ss and ff. In one query you can provide the set of nodes {a1,a2,…,ac}{a1,a2,…,ac} of the tree. As a result, you will get two numbers aiai and dist(ai,s)+dist(ai,f)dist(ai,s)+dist(ai,f). The node aiai is any node from the provided set, for which the number dist(ai,s)+dist(ai,f)dist(ai,s)+dist(ai,f) is minimal.

You can ask no more than 1111 queries.

Input

The first line contains a single integer tt (1≤t≤10)(1≤t≤10) — the number of test cases. Please note, how the interaction process is organized.

The first line of each test case consists of a single integer nn (2≤n≤1000)(2≤n≤1000) — the number of nodes in the tree.

The next n−1n−1 lines consist of two integers uu, vv (1≤u,v≤n,u≠v)(1≤u,v≤n,u≠v) — the edges of the tree.

Interaction

To ask a query print a single line:

  • In the beginning print "? c " (without quotes) where cc (1≤c≤n)(1≤c≤n) denotes the number of nodes being queried, followed by cc distinct integers in the range [1,n][1,n]  — the indices of nodes from the list.

For each query, you will receive two integers xx, dd — the node (among the queried nodes) with the minimum sum of distances to the hidden nodes and the sum of distances from that node to the hidden nodes. If the subset of nodes queried is invalid or you exceeded the number of queries then you will get x=d=−1x=d=−1. In this case, you should terminate the program immediately.

When you have guessed the hidden nodes, print a single line "! " (without quotes), followed by two integers in the range [1,n][1,n]  — the hidden nodes. You can output the hidden nodes in any order.

After this, you should read a string. If you guess the nodes correctly, you will receive the string "Correct". In this case, you should continue solving the remaining test cases or terminate the program, if all test cases were solved. Otherwise, you will receive the string "Incorrect". In this case, you should terminate the program immediately.

Guessing the hidden nodes does not count towards the number of queries asked.

The interactor is not adaptive. The hidden nodes do not change with queries.

Do not forget to read the string "Correct" / "Incorrect" after guessing the hidden nodes.

You need to solve each test case before receiving the input for the next test case.

The limit of 1111 queries applies to each test case and not to the entire input.

After printing a query do not forget to output the end of the line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  • fflush(stdout) or cout.flush() in C++;
  • System.out.flush() in Java;
  • flush(output) in Pascal;
  • stdout.flush() in Python;
  • see the documentation for other languages.

Hacks

To hack the solution, use the following test format:

The first line should contain a single integer tt (1≤t≤10)(1≤t≤10) — the number of test cases. The description of the test cases follows.

The first line of each test case should contain a single integer nn (2≤n≤1000)(2≤n≤1000) — the number of nodes in the tree. The second line should contain two distinct integers in the range [1,n][1,n]  — the hidden nodes. The next n−1n−1 lines should contain two integers uu, vv (1≤u,v≤n,u≠v)(1≤u,v≤n,u≠v)  — the edges of the tree.

Example

input

Copy

1
3
1 2
1 3

1 1

2 3

3 1

3 1

Correct

output

Copy

? 1 1

? 1 2

? 1 3

? 2 2 3

! 1 3

Note

The tree from the first test is shown below, and the hidden nodes are 11 and 33.

题意(互动题):

给你n(<=1e3),再给n-1条边构成树,你需要找到两个节点u和v。

每次询问给出n个点的下标,系统会回答你这n个点中,到u和v的距离之和最小的点x的下标(多个相同的随机给你其中一个),和距离之和。

你最多使用不超过11次询问,找出u和v。

思路:

刚看到这题的时候一脸懵逼,想了两天也没想出来。。。

实际上我们先查询一次所有的点,这样就可以得到一个位于路径(u,v)之间的点rt,和u到v的距离sum。

以rt为根,这样建立的树中,u和v的lca一定是rt。

这时我们只需要找出u,v其中一个点,然后以这个点为根再重新建树,查询一次距离这个点sum的所有节点,就可以得到另一个点。

核心问题:怎么找其中的一个点呢?

考虑使用二分。将所有的点先按照到rt的距离分类装进vector,二分距离询问对应的vector,根据得到的值是否大于sum来确定是否是位于u,v路径上离rt最远的一个点。

二分需要询问最多10次,刚开始1次,最后找另一个点又1次,这一共是12次。接下来就是最神奇的操作:

由于u,v路径上离rt最远的一个点,到rt的距离>=(sum+1)/2,即sum/2向上取整。因此我们二分时,下界直接从(sum+1)/2开始,减少一次二分,就减少了一次询问。

代码:

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define mst(head,x,n) memset(head+1,x,n*sizeof(head[0]))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dep(i,a,b) for(int i=(a);i>=(b);i--)
using namespace std;
const int maxn=1e3+5;
//const double pi=acos(-1.0);
//const double eps=1e-9;
//const ll mo=1e9+7;
int n,m,k;
int a[maxn],c[maxn];
int ans,tmp,cnt,rt,sum,ma;
int flag;
char s[maxn];
bool ok[maxn];
vector<int>vc[maxn],vv[maxn];
template <typename T>
inline void read(T &X){
    X=0;int w=0; char ch=0;
    while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
    while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    if(w) X=-X;
}
pair<int,int> query(int id){
    printf("? %d",vv[id].size());
    for(int i=0;i<vv[id].size();i++){
        printf(" %d",vv[id][i]);
    }
    puts("");
    fflush(stdout);
    int x,y;
    read(x);read(y);
    return make_pair(x,y);
}
void dfs(int u,int fa,int dep){
    ma=max(ma,dep);
    vv[dep].push_back(u);
    for(int i=0;i<vc[u].size();i++){
        int v=vc[u][i];
        if(v==fa) continue;
        dfs(v,u,dep+1);
    }
}
int jud(int as){
    pair<int,int>k=query(as);
    if(k.second>sum) return 1;
    rt=k.first;
    return 0;
}
int main(){
/*
#ifdef ONLINE_JUDGE
#else
    freopen("D:/Temp/in.txt", "r", stdin);
#endif
*/
    int T,cas=1;
    read(T);
    while(T--)
    {
        read(n);
        ans=0;
        rep(i,0,n+1) {vc[i].clear();vv[i].clear();}
        rep(i,1,n-1){
            int x,y;
            read(x);read(y);
            vc[x].push_back(y);
            vc[y].push_back(x);
            vv[n+1].push_back(i);
        }
        vv[n+1].push_back(n);
        pair<int,int>tp=query(n+1);
        rt=tp.first; sum=tp.second;
        ma=0;
        dfs(rt,-1,0);
        //cout<<"* "<<endl;
        ma=min(ma,sum);
        int l=(sum+1)/2,r=ma+1,tmp;
        while(l<r){
            int mid=(l+r)>>1;
            if(jud(mid)){
                r=mid;
            }
            else {l=mid+1;tmp=mid;}
        }
        rep(i,0,n) vv[i].clear();
        dfs(rt,-1,0);
        tp=query(sum);
        printf("! %d %d\n",rt,tp.first);
        fflush(stdout);
        scanf("%s",s);
    }
    return 0;
}
/*
1
3
1 2
1 3
1 1
*/

 

 

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