题目连接

http://poj.org/problem?id=3580

SuperMemo

Description

Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:

  1. ADD x y D: Add D to each number in sub-sequence {Ax ... Ay}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
  2. REVERSE x y: reverse the sub-sequence {Ax ... Ay}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
  3. REVOLVE x y T: rotate sub-sequence {Ax ... AyT times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
  4. INSERT x P: insert P after Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
  5. DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
  6. MIN x y: query the participant what is the minimum number in sub-sequence {Ax ... Ay}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2

To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.

Input

The first line contains (≤ 100000).

The following n lines describe the sequence.

Then follows M (≤ 100000), the numbers of operations and queries.

The following M lines describe the operations and queries.

Output

For each "MIN" query, output the correct answer.

Sample Input

6
1
7
8
5
2
9
8
ADD 1 3 2
INSERT 4 7
REVERSE 1 7
MIN 3 5
DEL 2
DEL 3
REVOLVE 1 5 3
MIN 1 3

Sample Output

5
3

伸展树模板题,几乎所有的操作都在里面了。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<set>
using std::set;
using std::min;
using std::pair;
using std::swap;
using std::vector;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1 << 17;
const int INF = ~0u >> 1;
int arr[N];
struct Node {
bool rev;
int dat, s, mval, delta;
Node *fa, *ch[2];
inline void set(int _dat, int _s, Node *p) {
rev = delta =0;
dat = mval = _dat, s = _s;
fa = ch[0] = ch[1] = p;
}
inline bool dir() const {
return this == fa->ch[0];
}
inline void link(Node *x, bool d) {
ch[d] = x, x->fa = this;
}
inline void push_up() {
s = ch[0]->s + ch[1]->s + 1;
mval = min(dat, min(ch[0]->mval, ch[1]->mval));
}
inline void update() {
if(!s) return;
rev ^= 1;
swap(ch[0], ch[1]);
}
inline void add(int v) {
if(!s) return;
delta += v;
mval += v;
dat += v;
}
inline void push_down() {
if(delta) {
ch[0]->add(delta);
ch[1]->add(delta);
delta = 0;
}
if(rev) {
ch[0]->update();
ch[1]->update();
rev ^= 1;
}
}
};
#define RRL root->ch[1]->ch[0]
struct SplayTree {
int top;
Node *root, *tail;
Node *null, stack[N], *pool[N >> 1];
inline void init(int n) {
top = 0, tail = &stack[0];
null = tail++;
null->set(INF, 0, NULL);
root = newNode(INF);
root->link(newNode(INF), 1);
Node *x = built(1, n);
root->ch[1]->link(x, 0);
root ->ch[1]->push_up();
root->push_up();
splay(x, null);
}
inline Node *newNode(int v) {
Node *p = !top ? tail++ : pool[--top];
p->set(v, 1, null);
return p;
}
inline Node *built(int l, int r) {
if(l > r) return null;
int mid = (l + r) >> 1;
Node *p = newNode(arr[mid]);
p->ch[0] = built(l, mid - 1);
if(p->ch[0]->s) p->ch[0]->fa = p;
p->ch[1] = built(mid + 1, r);
if(p->ch[1]->s) p->ch[1]->fa = p;
p->push_up();
return p;
}
inline void rotate(Node *&x, bool d) {
Node *y = x->fa;
y->push_down(), x->push_down();
y->ch[!d] = x->ch[d];
if(x->ch[d]->s) x->ch[d]->fa = y;
x->fa = y->fa;
if(y->fa->s) y->fa->ch[!y->dir()] = x;
x->ch[d] = y;
y->fa = x, y->push_up();
if(y == root) root = x;
}
inline void splay(Node *x, Node *f) {
for(;x->fa !=f ;x->push_down()) {
if(x->fa->fa == f) {
rotate(x, x->dir());
} else {
bool d = x->fa->dir();
if(d == x->dir()) rotate(x->fa, d), rotate(x, d);
else rotate(x, !d), rotate(x, d);
}
}
x->push_up();
}
inline Node *select(Node *x, int k) {
for(int t = 0; x->s; ) {
x->push_down();
t = x->ch[0]->s;
if(t == k) break;
if(k < t) x = x->ch[0];
else k -= t + 1, x = x->ch[1];
}
return x;
}
inline void get_range(int l, int r) {
splay(select(root, l - 1), null);
splay(select(root, r + 1), root);
}
inline void add(int l, int r, int v) {
get_range(l, r);
RRL->add(v);
splay(RRL, null);
}
inline void insert(int k, int v) {
splay(select(root, k), null);
splay(select(root, k + 1), root);
root->ch[1]->link(newNode(v), 0);
root->ch[1]->push_up();
root->push_up();
splay(RRL, null);
}
inline void reverse(int l, int r) {
get_range(l, r);
RRL->update();
}
inline void revolve(int l, int r, int k) {
int len = (r - l + 1);
k = (k % len + len) % len;
if(!k) return;
if(1 == k) {
erase(r);
insert(l - 1, pool[--top]->dat);
} else {
splay(select(root, r - k + 1), null);
splay(select(root, r + 1), root);
splay(select(root, l - 1), root);
splay(select(root, r), root->ch[1]);
Node *x = root->ch[0]->ch[1];
root->ch[0]->ch[1] = null;
root->ch[0]->push_up();
root->ch[1]->ch[0]->link(x, 1);
splay(x, null);
}
}
inline void erase(int k) {
splay(select(root, k), null);
Node *ret = root;
root = root->ch[1];
root->fa = null;
splay(select(root, 0), null);
root->ch[0] = ret->ch[0];
root->ch[0]->fa = root;
root->push_up();
pool[top++] = ret;
}
inline void query(int l, int r) {
get_range(l, r);
printf("%d\n", RRL->mval);
}
}spt;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
char buf[20];
int n, q, a, b, c;
while(~scanf("%d", &n)) {
rep(i, n) scanf("%d", &arr[i + 1]);
spt.init(n);
scanf("%d", &q);
while(q--) {
scanf("%s", buf);
if('D' == buf[2]) {
scanf("%d %d %d", &a, &b, &c);
spt.add(a, b, c);
} else if('S' == buf[2]) {
scanf("%d %d", &a, &b);
spt.insert(a, b);
} else if('V' == buf[2] && 'E' == buf[3]) {
scanf("%d %d", &a, &b);
spt.reverse(a, b);
} else if('V' == buf[2] && 'O' == buf[3]) {
scanf("%d %d %d", &a, &b, &c);
spt.revolve(a, b, c);
} else if('L' == buf[2]) {
scanf("%d", &a);
spt.erase(a);
} else if('N' == buf[2]) {
scanf("%d %d", &a, &b);
spt.query(a, b);
}
}
}
return 0;
}

poj 3580 SuperMemo的更多相关文章

  1. POJ 3580 - SuperMemo - [伸展树splay]

    题目链接:http://poj.org/problem?id=3580 Your friend, Jackson is invited to a TV show called SuperMemo in ...

  2. 平衡树(Splay):Splaytree POJ 3580 SuperMemo

    SuperMemo         Description Your friend, Jackson is invited to a TV show called SuperMemo in which ...

  3. POJ 3580 SuperMemo (splay tree)

    SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 6841   Accepted: 2268 Case Ti ...

  4. Splay树(多操作)——POJ 3580 SuperMemo

    相应POJ题目:点击打开链接 SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 11309   Accept ...

  5. POJ 3580 SuperMemo (FHQ_Treap)

    题意:让你维护一个序列,支持以下6种操作: ADD x y d: 第x个数到第y个数加d . REVERSE x y : 将区间[x,y]中的数翻转 . REVOLVE x y t :将区间[x,y] ...

  6. POJ 3580 SuperMemo 伸展树

    题意: 维护一个序列,支持如下几种操作: ADD x y D:将区间\([x,y]\)的数加上\(D\) REVERSE x y:翻转区间\([x,y]\) REVOLVE x y T:将区间\([x ...

  7. POJ 3580:SuperMemo(Splay)

    http://poj.org/problem?id=3580 题意:有6种操作,其中有两种之前没做过,就是Revolve操作和Min操作.Revolve一开始想着一个一个删一个一个插,觉得太暴力了,后 ...

  8. 【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$x$ ...

  9. 【POJ 3580】 SuperMemo

    [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 ...

随机推荐

  1. fastcgi(一)

    首先安装 fastcgi 开发包 ... #wget http://www.fastcgi.com/dist/fcgi-current.tar.gz #tar -zxvf fcgi-current.t ...

  2. Django基础篇之数据库选择及相关操作

    在djanjo框架中我们最常用的框架分别就是mysql和sqlit了,下面我们将分别讲述一下这俩种数据库的基础必备知识 mysql 一.利用命令创建(在终端上执行) 1.首先创建一个project项目 ...

  3. plsql自定义快捷键

    说明:如 输入   sf按空格 就变成   SELECT * FROM 输入  w空格  就变成 WHERE 可以帮助你快速的写语句,配置如下图

  4. 学习记录 java session保存用户登录

    <%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding= ...

  5. Code Sign error: No unexpired provisioning profiles found that contain any of the keychain's signing certificates

    最近离职了,刚好在离职之际有人叫我帮做个项目,简直了,没有mac电脑,没有真ji设备,简直了.接项目那哥们,暂且叫做J,大哥说我给你想办法,then,给借了个mac pro.刚拿到电脑真是喜出望外啊, ...

  6. jquery的end(),addBack()方法example

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  7. gulp.spritesmith修改px为rem单位

    移动端开发中,使用gulp.spritesmith进行小图sprite并生成样式,但由于spritesmith默认是以px为单位,所以就把插件的内容修改了下让生成rem单位并且能把background ...

  8. JS常用的设计模式(17)—— 状态模式

    状态模式主要可以用于这种场景 1 一个对象的行为取决于它的状态 2 一个操作中含有庞大的条件分支语句 回想下街头霸王的游戏. 隆有走动,攻击,防御,跌倒,跳跃等等多种状态,而这些状态之间既有联系又互相 ...

  9. IE样式兼容写法

    1.第一种写法 利用<!--[if lt IE 6/7/8/9/10/11]><![endif]-->,给每个html写一个class <!DOCTYPE html> ...

  10. 从Git仓库中恢复已删除的分支、文件或丢失的commit

    亲测可用 因为自己 commit 并且 push 后 因为冲突 提交不了,不小心做了 rebase 代码被 覆盖 用以下命令 还原: 查看所有日志 并记下 hash 值 git reflog 然后用: ...