在PTA关闭同步流会wa=-= 

L2-004 这是二叉搜索树吗?

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e5+9;
const int mod=9973;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

ll power(ll x,ll n)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*x;
        x=x*x;
        n>>=1;
    }
    return ans;
}

int n;
vector<int> ans,a;
//int a[maxn];
bool mirror=0;
void dfs(int root,int tail)
{
    if(root>tail) return ;
    int l=root+1;
    int r=tail;
    if(!mirror)
    {
        while(a[l]<a[root]&&l<=tail) l++;
        while(a[r]>=a[root]&&r>root) r--;
    }
    else
    {
        while(a[l]>=a[root]&&l<=tail) l++;
        while(a[r]<a[root]&&r>root) r--;
    }
    if(l-r!=1) return ;
    dfs(root+1,r);//左子树
    dfs(l,tail);//右
    ans.push_back(a[root]);
}
int main()
{
    //ios::sync_with_stdio(false);
    cin>>n;
    a.resize(n);
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    dfs(0,n-1);
    if(ans.size()!=n)
    {
        ans.clear();
        mirror=1;
        dfs(0,n-1);
    }
    if(ans.size()==n)
    {
        printf("YES\n");
        for(int i=0;i<ans.size();i++)
        {
            if(i!=0) cout<<" ";
            cout<<ans[i];
        }
    }
    else printf("NO");
    return 0;

}

 

相关文章: