distance constrain there show owin exists sat its 分支

題目描述

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

輸入

The input is given from Standard Input in the following format:
N
a1 a2 … aN

輸出

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

樣例輸入

5
3 2 2 3 3

樣例輸出

Possible

提示

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.

解析

代碼比較長，整體思路就是先排序，找到最長的那一路，然後確定這個長度為主幹（比如說最長的是n，那主幹長度是n+1），然後把其他的分支都可以插在中間，但是這些分路構成的長度不能長於主幹長度，因此這些分支的距離就是主幹中間的最長距離加一，就比如說樣例，主幹長度是3，因此有4個數為主幹，因此這四個數的最長距離為3、2、2、3，剩下的一個可以在中間的兩個二中任選一個加一，就是3、2、2、3、3。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int s[200];
int main()
{
 
    ios::sync_with_stdio(
 
false);
    int i,j,k,m,n;
    int ans=0,cnt=0;
    int flag=0;
    int flag1=0;
    cin>>n;
    for(i=1; i<=n; i++) cin>>s[i];
    sort(s+1,s+n+1);
    int gen=s[n],num=0;
    for(i=1; i<=n; i++)
        if(s[i]==gen) num++;
    if(num==1) cout<<"Impossible"<<endl;
    else
    {
        if(gen%2!=0)
        {
 
            k=(gen+1)/2;
            m=k;
            for(i=1;i<=n;i++)
            {
                if(s[i]==k)
                {
                    cnt++;
                }
                if(cnt>2)
                {
                    cout<<"Impossible"<<endl;
                    return 0;
                }
            }
            for(i=1; i<=n; i++)
            {
                if(s[i]<k)
                {
                    cout<<"Impossible"<<endl;
                    return 0;
                }
                if(m==s[i])
                {
                    ans++;
                }
                if(ans==2)
                {
                    ans=0;
                    m++;
                }
                if(m==(gen+1))
                {
                    cout<<"Possible"<<endl;
                    return 0;
                }
            }
            cout<<"Impossible"<<endl;
        }
        else
        {
            k=gen/2;
            m=k;
             for(i=1;i<=n;i++)
            {
                if(s[i]==k)
                {
                    cnt++;
                }
                if(cnt>1)
                {
                    cout<<"Impossible"<<endl;
                    return 0;
                }
            }
            for(i=1; i<=n; i++)
            {
                if(s[i]<k)
                {
                    cout<<"Impossible"<<endl;
                    return 0;
                }
 
                if(m==s[i])
                {
                    ans++;
                }
                if(ans==2&&flag)
                {
                    ans=0;
                    m++;
                }
                else if(ans==1&&!flag)
                {
                    ans=0;
                    m++;
                    flag=1;
                }
                if(m==gen+1)
                {
                    cout<<"Possible"<<endl;
                    return 0;
                }
            }
            cout<<"Impossible"<<endl;
        }
    }
 
    return 0;
}

View Code

Tree Restoring