[HDU4585]Shaolin

Problem

问你一个数的前驱和后继

Solution

Treap模板题

Notice

注意输出那个人的编号

Code

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100001;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int point = 0, pre, suf, root;
struct node
{
    int Val[N + 5], Level[N + 5], Size[N + 5], Son[2][N + 5], Label[N + 5];
    inline void up(int u)
    {
        Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + 1;
    }
    inline void Newnode(int &u, int v, int t)
    {
        u = ++point;
        Level[u] = rand(), Val[u] = v, Label[u] = t;
        Size[u] = 1, Son[0][u] = Son[1][u] = 0;
    }
    inline void Lturn(int &x)
    {
        int y = Son[1][x]; Son[1][x] = Son[0][y], Son[0][y] = x;
        Size[y] = Size[x]; up(x); x = y;
    }
    inline void Rturn(int &x)
    {
        int y = Son[0][x]; Son[0][x] = Son[1][y], Son[1][y] = x;
        Size[y] = Size[x]; up(x); x = y;
    }

    void Insert(int &u, int t, int tt)
    {
        if (u == 0)
        {
            Newnode(u, t, tt);
            return;
        }
        Size[u]++;
        if (t < Val[u])
        {
            Insert(Son[0][u], t, tt);
            if (Level[Son[0][u]] < Level[u]) Rturn(u);
        }
        else if (t > Val[u])
        {
            Insert(Son[1][u], t, tt);
            if (Level[Son[1][u]] < Level[u]) Lturn(u);
        }
    }

    int Find_num(int u, int t)
    {
        if (!u) return 0;
        if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
        else if (t <= Size[Son[0][u]] + 1) return u;
        else return Find_num(Son[1][u], t - Size[Son[0][u]] - 1);
    }
    void Find_pre(int u, int t)
    {
        if (!u) return;
        if (t > Val[u])
        {
            pre = u;
            Find_pre(Son[1][u], t);
        }
        else Find_pre(Son[0][u], t);
    }
    void Find_suf(int u, int t)
    {
        if (!u) return;
        if (t < Val[u])
        {
            suf = u;
            Find_suf(Son[0][u], t);
        }
        else Find_suf(Son[1][u], t);
    }
}Treap;
int sqz()
{
    int n;
    while (~scanf("%d", &n) && n)
    {
        root = point = 0;
        int x = read(), y = read();
        printf("%d 1
", x);
        Treap.Insert(root, y, x);
        rep(i, 2, n)
        {
            pre = suf = -1;
            x = read(), y = read();
            Treap.Find_pre(root, y);
            Treap.Find_suf(root, y);
            Treap.Insert(root, y, x);
            printf("%d ", x);
            if (pre == -1) printf("%d
", Treap.Label[suf]);
            else if (suf == -1) printf("%d
", Treap.Label[pre]);
            else if (y - Treap.Val[pre] <= Treap.Val[suf] - y) printf("%d
", Treap.Label[pre]);
            else printf("%d
", Treap.Label[suf]);
        }
    }
}
原文地址:https://www.cnblogs.com/WizardCowboy/p/7643691.html