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Tree Restoring

2024-08-15 12:08:40   219人阅读

Tree Restoring

Time limit : 2sec / Memory limit : 256MB

Score : 700 points

Problem Statement

Aoki loves numerical sequences and trees.

One day, Takahashi gave him an integer sequence of length N, a1,a2,…,aN, which made him want to construct a tree.

Aoki wants to construct a tree with N vertices numbered 1 through N, such that for each i=1,2,…,N, the distance between vertex i and the farthest vertex from it is ai, assuming that the length of each edge is 1.

Determine whether such a tree exists.

Constraints

  • 2≦N≦100
  • 1≦ai≦N−1

Input

The input is given from Standard Input in the following format:

Na1 a2 … aN

Output

If there exists a tree that satisfies the condition, print Possible. Otherwise, print Impossible.

Sample Input 1

53 2 2 3 3

Sample Output 1

Possible

技术分享

The diagram above shows an example of a tree that satisfies the conditions. The red arrows show paths from each vertex to the farthest vertex from it.

分析：对于一棵树来说，假设直径有两个端点a，b，那么任意一点到其他点最远距离必然是max(dist(p,a),dist(p,b))，

　　　那么根据直径来构树，以树直径为奇数举例，那么这条链上必然有偶数个点，且最远距离为k,k-1,...,k/2,k/2...,k-1,k；

　　　那么也就是不存在最远距小于k/2的点，且k/2有两个点，大于k/2的至少有2个；

　　　树直径为偶数时同理；

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=1e5+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m,k,t;inline ll read(){    ll x=0;int f=1;char ch=getchar();    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}    return x*f;}int a[maxn],ma,vis[maxn];bool flag;int main(){    int i,j;    scanf("%d",&n);    rep(i,1,n)scanf("%d",&a[i]),vis[a[i]]++,ma=max(ma,a[i]);    if(ma%2==0)    {        rep(i,1,ma/2-1)if(vis[i])flag=true;        rep(i,ma/2,ma)        {            if(i==ma/2)            {                if(vis[i]!=1)flag=true;            }            else if(vis[i]<2)flag=true;        }    }    else    {        rep(i,1,(ma+1)/2-1)if(vis[i])flag=true;        rep(i,(ma+1)/2,ma)        {            if(i<(ma+1)/2&&vis[i])flag=true;            if(i==(ma+1)/2)            {                if(vis[i]!=2)flag=true;            }            else if(vis[i]<2)flag=true;        }    }    if(flag)puts("Impossible");    else puts("Possible");    //system("Pause");    return 0;}

Tree Restoring


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