poj 3580 SuperMemo
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, {A1, A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:
- 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}
- 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}
- REVOLVE x y T: rotate sub-sequence {Ax ... Ay} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
- 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}
- DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
- 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 n (n ≤ 100000).
The following n lines describe the sequence.
Then follows M (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
- 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 ...
- 平衡树(Splay):Splaytree POJ 3580 SuperMemo
SuperMemo Description Your friend, Jackson is invited to a TV show called SuperMemo in which ...
- POJ 3580 SuperMemo (splay tree)
SuperMemo Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 6841 Accepted: 2268 Case Ti ...
- 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 ...
- 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] ...
- 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 ...
- 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 ...
- 【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$x$ ...
- 【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
- 前台页面验证中需要注意的一个与VARCHAR2(N BYTE)和VARCHAR2(N CHAR)的小细节
1:一个小的测试实例 CREATE TABLE SALES.TEST_ ( TEST_BYTE BYTE), TEST_CHAR CHAR) )--TABLE CREATED ')--1 ROW IN ...
- devexpress中ASPxGridView控件初始化赋值
写在ASPxGridView中OnCellEditorInitialize="ASPxGridView_progoods_CellEditorInitialize" 事件中: / ...
- WWF3状态机工作流<WWF第七篇>
状态机是另外一种常见的工作流类型.它是以状态的变迁为驱动而进行业务流转的,是一定需要人为干预的,而不像顺序类型工作流那样可以按照事先设计好的业务流程一步一步依次执行下去. 一.状态机工作流范例 Sta ...
- css经验点滴积累
1.filter:alpha(opacity=70);-moz-opacity:0.7;-webkit-opacity: 0.7;-o-opacity: 0.7;-ms-opacity: 0.7;op ...
- linux 文本处理
tr,awk,sed 一:tr 1.大小写转换 cat file | tr [a-z] [A-Z] > new_file(大写 --> 小写) cat file | tr [A-Z] [a ...
- Linux之磁盘管理
本章重点提示: 1):理解基础命令,df,fdisk. 2):磁盘分区的理论基础. 1:查看当前系统分区与挂载情况: [root@localhost ~]# df Filesystem 1K-bloc ...
- minicom/kermit捕捉日志
1.minicom捕捉日志 ctrl-A Z 命令窗口中有 Capture on/off......L 2.kermit捕捉日志 ctrl-\ C进入kermit命令行模式 log session ...
- Html5元素及基本语法
HTML标签开始标签(opening tag):开放标签结束标签(closing tag):闭合标签 元素定义:HTML元素指的是从开始标签到结束标签的代码(元素以开始标签为起始以借宿标签终止)元素的 ...
- kettle的hello world
本篇介绍使用kettle的一个最简单的例子,可以初步了解下转换. 需求是这样的: 存在一个本地csv文件,文件的内容如下 现在需要将csv中的数据保存到本地的文本文件中 1.创建一个转换,并且重命名 ...
- 当数据0跟if判断冲突的时候
我是很无奈的,以后都要2,3,4,5这样去标志状态: 分配状态:<select name="is_send" > <option selected="s ...