• [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]);
            }
        }
    }
    
  • 相关阅读:
    Linkedin工程师是如何优化他们的Java代码的
    如何调试 Android 上 HTTP(S) 流量
    Facebook工程师是如何改进他们Android客户端的
    modified: xxx(modified content, untracked content)
    在Android工程中加入AIDL文件时,gen目录生成的文件报错-问题解决
    Ubuntu14.04LTS下使用eclipse搭建Cocos2d-x的Android环境
    Ubuntu Linux下安装Oracle JDK
    查看Android源码版本
    Ubuntu Linux下安装Oracle JDK
    Android Initializing a Build Environment
  • 原文地址:https://www.cnblogs.com/WizardCowboy/p/7643691.html
Copyright © 2020-2023  润新知