Problem Statement

This is an interactive task.

Let N be an odd number at least 3.

There are N seats arranged in a circle. The seats are numbered 0 through N−1. For each i (0≤i≤N−2), Seat i and Seat i+1are adjacent. Also, Seat N−1 and Seat 0 are adjacent.

Each seat is either vacant, or oppupied by a man or a woman. However, no two adjacent seats are occupied by two people of the same sex. It can be shown that there is at least one empty seat where N is an odd number at least 3.

You are given N, but the states of the seats are not given. Your objective is to correctly guess the ID number of any one of the empty seats. To do so, you can repeatedly send the following query:

• Choose an integer i (0≤i≤N−1). If Seat i is empty, the problem is solved. Otherwise, you are notified of the sex of the person in Seat i.

Guess the ID number of an empty seat by sending at most 20 queries.

Constraints

• N is an odd number.
• 3≤N≤99 999

Input and Output

First, N is given from Standard Input in the following format:

N

Then, you should send queries. A query should be printed to Standart Output in the following format. Print a newline at the end.

i

The response to the query is given from Standard Input in the following format:

s

Here, s is Vacant, Male or Female. Each of these means that Seat i is empty, occupied by a man and occupied by a woman, respectively.

Notice

• Flush Standard Output each time you print something. Failure to do so may result in TLE.
• Immediately terminate the program when s is Vacant. Otherwise, the verdict is indeterminate.
• The verdict is indeterminate if more than 20 queries or ill-formatted queries are sent.

Sample Input / Output 1

In this sample, N=3, and Seat 0, 1, 2 are occupied by a man, occupied by a woman and vacant, respectively.

Problem Statement

You are given a forest with N vertices and M edges. The vertices are numbered 0 through N−1. The edges are given in the format (xi,yi), which means that Vertex xi and yi are connected by an edge.

Each vertex i has a value ai. You want to add edges in the given forest so that the forest becomes connected. To add an edge, you choose two different vertices i and j, then span an edge between i and j. This operation costs ai+aj dollars, and afterward neither Vertex i nor j can be selected again.

Find the minimum total cost required to make the forest connected, or print Impossible if it is impossible.

Constraints

• 1≤N≤100,000
• 0≤M≤N−1
• 1≤ai≤109
• 0≤xi,yi≤N−1
• The given graph is a forest.
• All input values are integers.

Sample Input 1

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

7

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1+6=7 dollars.

Sample Input 2

5 0
3 1 4 1 5

Sample Output 2

Impossible

We can't make the graph connected.

Sample Input 3

1 0
5

Sample Output 3

0

The graph is already connected, so we do not need to add any edges.

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
int a[N],fa[N],vis[N],mi[N];
pair<int,int>t;
priority_queue<pair<int,int>, vector<pair<int,int> >, greater<pair<int,int> > >Q;
int find(int x)
{return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{ios::sync_with_stdio(false);int n,m;cin>>n>>m;for(int i=0; i<n; i++)cin>>a[i],fa[i]=i,mi[i]=INF;for(int i=0,u,v; i<m; i++){cin>>u>>v,u=find(u),v=find(v);if(u!=v)fa[u]=v;}for(int i=0; i<n; i++)find(i);for(int i=0; i<n; i++)mi[fa[i]]=min(a[i],mi[fa[i]]),Q.push(make_pair(a[i],fa[i]));sort(fa,fa+n);int tn=unique(fa,fa+n)-fa;if(tn==1)cout<<0;else{long long s=0;for(int i=0; i<tn; i++)s+=mi[fa[i]];while(!Q.empty()){if(tn==2)break;t=Q.top();Q.pop();if(!vis[t.second]){vis[t.second]=1;continue;}s+=t.first,--tn;}if(tn==2)cout<<s;else cout<<"Impossible";}return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
int a[N],fa[N],vis[N],mi[N];
typedef pair<int,int> pii;
priority_queue<pii, vector<pii >, greater<pii > >Q;
int find(int x)
{return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{ios::sync_with_stdio(false);int n,m;cin>>n>>m;for(int i=0; i<n; i++)cin>>a[i],fa[i]=i,mi[i]=INF;for(int i=0,u,v; i<m; i++){cin>>u>>v,u=find(u),v=find(v);if(u!=v)fa[u]=v;}for(int i=0; i<n; i++)find(i);for(int i=0; i<n; i++)mi[fa[i]]=min(a[i],mi[fa[i]]),Q.push(make_pair(a[i],fa[i]));sort(fa,fa+n);int tn=unique(fa,fa+n)-fa;if(tn==1)cout<<0;else if((tn-1)*2>n)cout<<"Impossible";else{long long s=0;for(int i=0; i<tn; i++)s+=mi[fa[i]];while(!Q.empty()){if(tn==2)break;pii t=Q.top();Q.pop();if(!vis[t.second]){vis[t.second]=1;continue;}s+=t.first,--tn;}cout<<s;}return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
typedef pair<int,int> pii;
int fa[N],vis[N],mi[N],V[N];
vector<pii >Q(N);
int find(int x)
{return x==fa[x]?x:(fa[x]=find(fa[x]));
}
int main()
{ios::sync_with_stdio(false);int n,m,tn=0;cin>>n>>m;for(int i=0; i<n; i++)cin>>Q[i].first,fa[i]=i,mi[i]=INF;for(int i=0,u,v; i<m; i++){cin>>u>>v,u=find(u),v=find(v);if(u!=v)fa[u]=v;}for(int i=0; i<n; i++){find(i);if(!vis[fa[i]])V[tn++]=fa[i];vis[fa[i]]=1,Q[i].second=fa[i],mi[fa[i]]=min(Q[i].first,mi[fa[i]]);}if(tn==1)cout<<0;else if((tn-1)*2>n)cout<<"Impossible";else{long long s=0;for(int i=0; i<tn; i++)s+=mi[V[i]];if(tn!=2)sort(Q.begin(),Q.begin()+n);for(int i=0; i<n&&tn!=2; i++){if(vis[Q[i].second]){vis[Q[i].second]=0;continue;}s+=Q[i].first,tn--;}cout<<s;}return 0;
}

Score : 900 points

N
a0 b0
a1 b1
:
aN−2 bN−2

5
0 1
0 2
0 3
3 4

2

2
0 1

1

10
2 8
6 0
4 1
7 6
2 3
8 6
6 9
2 4
5 8

3

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
vector<int>G[N];
int n,dp[N],f=-1;
void dfs(int i,int p)
{int F=0;for(auto j:G[i]){if(j==p)continue;dfs(j,i);dp[i]+=dp[j];if(!dp[j]){if(!F)F=1;else dp[i]++;}}
}
int main()
{scanf("%d",&n);for(int i=1,u,v; i<n; i++)scanf("%d%d",&u,&v),G[u].push_back(v),G[v].push_back(u);for(int i=0; i<n; i++)if(G[i].size()>2){f=i;break;}if(f==-1)printf("1\n");else dfs(f,-1),printf("%d\n",dp[f]);return 0;
}