hdu 4585 Shaolin
原题链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=46093
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
const int Max_N = ;
struct Node {
int v, s;
Node *ch[];
inline void set(int _v, int _s , Node *p) {
ch[] = ch[] = p;
v = _v, s = _s;
}
inline void push_up() {
s = ch[]->s + ch[]->s + ;
}
inline int cmp(int x) const {
return x > v;
}
};
struct SizeBalanceTree {
Node *tail, *root, *null;
Node stack[Max_N];
void init(){
tail = &stack[];
null = tail++;
null->set(, , NULL);
root = null;
}
inline Node *newNode(int v) {
Node *p = tail++;
p->set(v, , null);
return p;
}
inline void rotate(Node* &x, int d) {
Node *k = x->ch[!d];
x->ch[!d] = k->ch[d];
k->ch[d] = x;
k->s = x->s;
x->push_up();
x = k;
}
inline void Maintain(Node* &x, int d) {
if (x->ch[d] == null) return;
if (x->ch[d]->ch[d]->s > x->ch[!d]->s) {
rotate(x, !d);
} else if (x->ch[d]->ch[!d]->s > x->ch[!d]->s) {
rotate(x->ch[d], d), rotate(x, !d);
} else {
return;
}
Maintain(x, ), Maintain(x, );
}
inline void insert(Node* &x, int v) {
if (x == null) {
x = newNode(v);
return;
} else {
x->s++;
int d = x->cmp(v);
insert(x->ch[d], v);
x->push_up();
Maintain(x, d);
}
}
inline void insert(int v) {
insert(root, v);
}
inline int count(int v) {
Node *x = root;
int res = , t = ;
for (; x->s;) {
t = x->ch[]->s;
if (v < x->v) x = x->ch[];
else res += t + , x = x->ch[];
}
return res;
}
inline int kth(int k) {
int t = ;
Node *x = root;
if (x->s < k) return -;
for (; x->s;) {
t = x->ch[]->s;
if (k == t + ) break;
else if (k <= t) x = x->ch[];
else k -= t + , x = x->ch[];
}
return x->v;
}
inline int operator[](int k) {
return kth(k);
}
}sbt;
int id[];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int n, t, x, a, b, k;
while (~scanf("%d", &n) && n) {
sbt.init();
scanf("%d%d", &t, &x);
sbt.insert(x), id[x] = t;
printf("%d 1\n", t);
for (int i = ; i <= n; i++) {
scanf("%d%d", &t, &x);
id[x] = t, k = sbt.count(x);
if (!k) a = sbt[];
else if (k == i - ) a = sbt[i - ];
else {
a = sbt[k];
b = sbt[k + ];
if (x - a > b - x) a = b;
}
printf("%d %d\n", t, id[a]);
sbt.insert(x);
}
}
return ;
}
hdu 4585 Shaolin的更多相关文章
- HDU 4585 Shaolin(Treap找前驱和后继)
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Su ...
- HDU 4585 Shaolin(STL map)
Shaolin Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit cid= ...
- [HDU 4585] Shaolin (map应用)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4585 题目大意:不停的插入数字,问你跟他相距近的ID号.如果有两个距离相近的话选择小的那个. 用map ...
- hdu 4585 Shaolin treap
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Problem ...
- A -- HDU 4585 Shaolin
Shaolin Time Limit: 1000 MS Memory Limit: 32768 KB 64-bit integer IO format: %I64d , %I64u Java clas ...
- hdu 4585 Shaolin(STL map)
Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shao ...
- HDU 4585 Shaolin (STL)
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Sub ...
- HDU 4585 Shaolin(水题,STL)
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Sub ...
- HDU 4585 Shaolin (STL map)
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Sub ...
随机推荐
- win8或win8.1修改注册表失败的原因
win8 and win8.1 modify the registry need compiled to be different versions according to the os bits.
- EXT学习之——Extjs 文本框 TextField 添加点击(onclick)事件方法
{ xtype:'textfield', listeners: { render: function(p) { // Append the Panel to the click handler's a ...
- python3的文件操作
open的原型定义在bultin.py中,是一种内建函数,用于处理文件 open(file, mode='r', buffering=None, encoding=None, errors=None, ...
- 【MVC】自定义ASP.NET MVC Html辅助方法
在ASP.NET MVC中,Html辅助方法给我们程序员带来很多方便,其重要性也就不言自明.有时候,我们不想重复地写一些HTML代码,或者MS没有提供我们想要的那个HTML标签的Html辅助方法,那么 ...
- HTTP MIME类型即HttpResponse.ContentType属性值列表
MIME-Typ Dateiendung(en) Bedeutung application/acad *.dwg AutoCAD-Dateien (nach NCSA) application/ap ...
- QQ聊天信息提取
先前在iOS 8.x版时,往往未能顺利取出QQ的聊天信息,即使顺利取出数据库,却发现聊天信息已被加密处理,仅只能得知是与哪些QQ号进行聊天,而未能顺利得知聊天内容. 但这个情况到后来有了变化,以下情境 ...
- Android代码写View
1.Java Code package com.fish.helloworld; import android.app.Activity; import android.content.Context ...
- RedHat安装VMwareTools出现解压压缩包时无法打开文件的现象
出现这种情况的原因是因为解压命令没有加—C参数,使用的命令为:tat -xvzf VMware Tools: 正确的解压命令应该是: tar -xvzf VMware Tools -C /opt,加上 ...
- javaSE第二十四天
第二十四天 363 1:多线程(理解) 363 (1)JDK5以后的Lock锁 363 A:定义 363 B:方法: 364 C:具体应用(以售票程序为例) 364 ...
- app开发版面设计原则
(1) 单纯:形象和色彩必须简单明了(也就是简洁性). (2) 统一:造型与色彩必须和谐,要具有统一的协调效果. (3) 均衡:整个画面须要具有魄力感与均衡效果. (4) 展现重点:构成要素必须化繁为 ...