【题目链接】

点击打开链接

【算法】

本题也是Splay区间操作的模板题,不过要比BZOJ 3223要稍微复杂一些,做完此题后,我终于对Splay有了更深入的理解,有“拨开云雾见青天”的感觉

本题还是有许多细节的,笔者花了5h才通过了此题

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 100000
const int INF = 2e9; int i,N,M,d,P,x,y,t;
int a[MAXN+];
string opt; template <typename T> inline void read(T &x) {
int f=; x=;
char c = getchar();
for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
for (; isdigit(c); c = getchar()) x = x * + c - '';
x *= f;
} template <typename T> inline void write(T x) {
if (x < ) { putchar('-'); x = -x; }
if (x > ) write(x/);
putchar(x%+'');
} template <typename T> inline void writeln(T x) {
write(x);
puts("");
} struct Splay {
int root,total;
struct Node {
int fa,son[],size,add,Min,val;
bool rev;
} Tree[MAXN*+];
inline bool get(int x) {
return Tree[Tree[x].fa].son[] == x;
}
inline void build(int index,int l,int r) {
int mid = (l + r) >> ;
Tree[index].size = ;
Tree[index].add = Tree[index].rev = ;
Tree[index].val = Tree[index].Min = a[mid];
if (l == r) return;
if (l < mid) {
++total;
Tree[index].son[] = total;
Tree[total].fa = index;
build(total,l,mid-);
Tree[index].size += Tree[Tree[index].son[]].size;
Tree[index].Min = min(Tree[index].Min,Tree[Tree[index].son[]].Min);
}
if (mid < r) {
++total;
Tree[index].son[] = total;
Tree[total].fa = index;
build(total,mid+,r);
Tree[index].size += Tree[Tree[index].son[]].size;
Tree[index].Min = min(Tree[index].Min,Tree[Tree[index].son[]].Min);
}
}
inline void new_node(int index,int x,int f) {
Tree[index].rev = ;
Tree[index].size = ;
Tree[index].val = Tree[index].Min = x;
Tree[index].add = ;
Tree[index].fa = f;
Tree[index].son[] = Tree[index].son[] = ;
}
inline int query_pos(int x) {
int index = root;
while (true) {
pushdown(index);
if (x > Tree[Tree[index].son[]].size) {
x -= Tree[Tree[index].son[]].size;
if (x == ) return index;
--x;
index = Tree[index].son[];
} else index = Tree[index].son[];
}
}
inline void pushdown(int index) {
if (Tree[index].rev) {
swap(Tree[index].son[],Tree[index].son[]);
Tree[Tree[index].son[]].rev ^= ;
Tree[Tree[index].son[]].rev ^= ;
Tree[index].rev = ;
}
if (Tree[index].add) {
Tree[Tree[index].son[]].val += Tree[index].add;
Tree[Tree[index].son[]].val += Tree[index].add;
Tree[Tree[index].son[]].add += Tree[index].add;
Tree[Tree[index].son[]].add += Tree[index].add;
Tree[Tree[index].son[]].Min += Tree[index].add;
Tree[Tree[index].son[]].Min += Tree[index].add;
Tree[index].add = ;
}
}
inline void update(int index) {
Tree[index].size = Tree[Tree[index].son[]].size + Tree[Tree[index].son[]].size + ;
Tree[index].Min = Tree[index].val;
if (Tree[index].son[]) Tree[index].Min = min(Tree[index].Min,Tree[Tree[index].son[]].Min);
if (Tree[index].son[]) Tree[index].Min = min(Tree[index].Min,Tree[Tree[index].son[]].Min);
}
inline void splay(int x,int pos) {
int f;
for (f = Tree[x].fa; (f = Tree[x].fa) != pos; rotate(x)) {
if (Tree[f].fa != pos)
rotate(get(f) == get(x) ? (f) : (x));
}
if (!pos) root = x;
}
inline void rotate(int x) {
int f = Tree[x].fa,g = Tree[f].fa,
tmpx = get(x),tmpf = get(f);
pushdown(f); pushdown(x);
if (!f) return;
Tree[f].son[tmpx] = Tree[x].son[tmpx^];
if (Tree[x].son[tmpx^]) Tree[Tree[x].son[tmpx^]].fa = f;
Tree[x].son[tmpx^] = f;
Tree[f].fa = x;
Tree[x].fa = g;
if (g) Tree[g].son[tmpf] = x;
update(f);
update(x);
}
inline void Insert(int p,int val) {
int x = query_pos(p),
y = query_pos(p+);
splay(x,); splay(y,root);
new_node(++total,val,Tree[root].son[]);
Tree[Tree[root].son[]].son[] = total;
update(Tree[root].son[]);
update(root);
}
inline void erase(int p) {
int x = query_pos(p-),
y = query_pos(p+);
splay(x,); splay(y,root);
Tree[Tree[root].son[]].son[] = ;
update(Tree[root].son[]);
update(root);
}
inline void add(int l,int r,int v) {
int x = query_pos(l-),
y = query_pos(r+);
splay(x,); splay(y,root);
Tree[Tree[Tree[root].son[]].son[]].val += v;
Tree[Tree[Tree[root].son[]].son[]].add += v;
Tree[Tree[Tree[root].son[]].son[]].Min += v;
update(Tree[root].son[]);
update(root);
}
inline void reverse(int l,int r) {
int x = query_pos(l-),
y = query_pos(r+);
splay(x,); splay(y,root);
Tree[Tree[Tree[root].son[]].son[]].rev ^= ;
}
inline int query_min(int l,int r) {
int x = query_pos(l-),
y = query_pos(r+);
splay(x,); splay(y,root);
return Tree[Tree[Tree[root].son[]].son[]].Min;
}
inline void revolve(int l,int r,int t) {
int x = query_pos(r-t),
y = query_pos(r+);
splay(x,); splay(y,root);
int tmp = Tree[Tree[root].son[]].son[];
Tree[Tree[root].son[]].son[] = ;
update(Tree[root].son[]);
update(root);
x = query_pos(l-);
y = query_pos(l);
splay(x,); splay(y,root);
Tree[Tree[root].son[]].son[] = tmp;
Tree[tmp].fa = Tree[root].son[];
update(Tree[root].son[]);
update(root);
}
} T; int main() { read(N); T.root = T.total = ;
for (i = ; i <= N + ; i++) read(a[i]);
a[] = a[N+] = INF;
T.build(,,N+); read(M);
while (M--) {
cin >> opt;
if (opt[] == 'A') {
read(x); read(y); read(d);
if (x > y) swap(x,y);
T.add(x+,y+,d);
} else if (opt[] == 'R' && opt[] == 'E' && opt[] == 'V' && opt[] == 'E'){
read(x); read(y);
if (x > y) swap(x,y);
T.reverse(x+,y+);
} else if (opt[] == 'R' && opt[] == 'E' && opt[] == 'V' && opt[] == 'O') {
read(x); read(y); read(t);
if (x > y) swap(x,y);
t = (t % (y - x + ) + y - x + ) % (y - x + );
if (!t) continue;
T.revolve(x+,y+,t);
} else if (opt[] == 'I') {
read(x); read(P);
T.Insert(x+,P);
} else if (opt[] == 'D') {
read(x);
T.erase(x+);
} else if (opt[] == 'M') {
read(x); read(y);
if (x > y) swap(x,y);
writeln(T.query_min(x+,y+));
}
} return ; }

【POJ 3580】 SuperMemo的更多相关文章

  1. 【POJ 3580】SuperMemo Splay

    题意 给定$n$个数,$m$个询问,每次在$[L,R]$区间加上一个数,或者反转一个区间$[L,R]$,或者循环右移区间$[L,R]$共$T$次,或者在第$x$个数后插入一个数$p$,或者删除第$x$ ...

  2. bzoj 2295: 【POJ Challenge】我爱你啊

    2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec  Memory Limit: 128 MB Description ftiasch是个十分受女生欢迎的同学,所以 ...

  3. 【链表】BZOJ 2288: 【POJ Challenge】生日礼物

    2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 382  Solved: 111[Submit][S ...

  4. BZOJ2288: 【POJ Challenge】生日礼物

    2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 284  Solved: 82[Submit][St ...

  5. BZOJ2293: 【POJ Challenge】吉他英雄

    2293: [POJ Challenge]吉他英雄 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 80  Solved: 59[Submit][Stat ...

  6. BZOJ2287: 【POJ Challenge】消失之物

    2287: [POJ Challenge]消失之物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 254  Solved: 140[Submit][S ...

  7. BZOJ2295: 【POJ Challenge】我爱你啊

    2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 126  Solved: 90[Submit][Sta ...

  8. BZOJ2296: 【POJ Challenge】随机种子

    2296: [POJ Challenge]随机种子 Time Limit: 1 Sec  Memory Limit: 128 MBSec  Special JudgeSubmit: 114  Solv ...

  9. BZOJ2292: 【POJ Challenge 】永远挑战

    2292: [POJ Challenge ]永远挑战 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 513  Solved: 201[Submit][ ...

随机推荐

  1. 使用Crypto对数据进行加密解密

    注释都在代码里: 先撸客户端: from Crypto.Cipher import AES import base64,requests class Message(object): def __in ...

  2. Maven自动部署(SCM-SVN/Git)(maven-scm-plugin/maven-release-plugin插件的使用)

    以下内容引用自https://ayayui.gitbooks.io/tutorialspoint-maven/content/book/maven_deployment_automation.html ...

  3. Limitations of Forms Personalization (文档 ID 420518.1)

    In this Document   Purpose   Scope   Details   Diagnostics & Utilities Community:   References A ...

  4. Android ZXing 二维码、条形码扫描介绍

    本帖最后由 Shims 于 2013-11-9 12:39 编辑 最近公司的Android项目需要用到摄像头做条码或二维码的扫描,Google一下,发现一个开源的 ZXing项目.它提供二维码和条形码 ...

  5. hdoj-1856-More is better【并查集】

    More is better Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others) To ...

  6. python中pymysql使用

    python中pymysql使用 https://blog.csdn.net/johline/article/details/69549131 import pymysql # 连接数据库 conne ...

  7. mahout in Action2.2-聚类介绍-K-means聚类算法

    聚类介绍 本章包含 1 实战操作了解聚类 2.了解相似性概念 3 使用mahout执行一个简单的聚类实例 4.用于聚类的各种不同的距离測算方法 作为人类,我们倾向于与志同道合的人合作-"鸟的 ...

  8. 第04章-VTK基础(2)

    [译者:这个系列教程是以Kitware公司出版的<VTK User's Guide -11th edition>一书作的中文翻译(出版时间2010年,ISBN: 978-1-930934- ...

  9. javascript读取和改动原型特别须要注意的事儿,由于原型的读写不具有对等性

    对于从原型对象继承而来的成员,其读和写具有内在的不正确等性.比方有一个对象A,假设它的原型对象是B.B的原型对象是null.假设我们须要读取A对象的name属性值,那么JS会优先在A中查找.假设找到了 ...

  10. 2016-1-8 windows 7下安装mysql及其配置和运用

    绪言 最近学习了一下mysql的相关用法,以及vs2010结合mysql的使用. 遇到的问题:1.安装mysql 5.6 绿色免安装版本,出现mysql server not connect loca ...