【POJ 3580】 SuperMemo
【题目链接】
【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
- 【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$x$ ...
- 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 ...
- 【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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- commons.apache
1.ToStringBuilder //对象及其属性一行显示 System.out.println(ToStringBuilder.reflectionToString(u)); System.out ...
- centos6安装概述
1.1.选择安装类型:[Install or upgrade an existing system]安装或升级现有系统 1.2.介质校验:[Skip]跳过介质校验,校验时间较长 1.3.语言选择:[E ...
- Java屏幕截图工具 捕获屏幕
原文:http://www.open-open.com/code/view/1420037709781 import java.awt.BorderLayout; import java.awt.Co ...
- Oracle版本–EBS R12.1.1
select * from v$version; Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - 64bit Productio ...
- scss - 语法
1.在一个样式导入另一个样式(@import "example.css") 2.scss单行注释不会显示出来 3.强大的变量(定义后,全局可使用) 4.全局默认变量(加!defau ...
- python实现QQ机器人(自己主动登录,获取群消息,发送群消息)
一次偶然的机会我看见了一个群里的一个QQ号总是依据你所发的消息自己主动回复,当时非常感觉到奇妙.我知道能够模拟登录站点,没想到居然也能模拟登录QQ,首先自己想到的就是怎样实现模拟登录PC端的QQ, 開 ...
- Effective C++ 条款13/14 以对象管理资源 || 在资源管理类中小心拷贝行为
三.资源管理 资源就是一旦你使用了它,将来不用的时候必须归还系统.C++中最常用的资源就是动态内存分配.其实,资源还有 文件描述符.互斥器.图形界面中的字形.画刷.数据库连接.socket ...
- Intel MIC
http://en.wikipedia.org/wiki/Intel_MIC Intel MIC From Wikipedia, the free encyclopedia Intel Man ...
- this、call和apply、bind
this关键字: JavaScript的this关键字,总是指向一个对象,具体指向哪个对象,是根据运行时函数指向环境动态绑定的.简单来说,this就是谁调用指向谁.具体使用中,this的指向,大致可以 ...
- 从.Net版本演变看String和StringBuilder性能之争
在C#中string关键字的映射实际上指向.NET基类System.String.System.String是一个功能非常强大且用途非常广泛的基类,所以我们在用C#string的时候实际就是在用.NE ...