SuperMemo
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12795 | Accepted: 3989 | |
| Case Time Limit: 2000MS | ||
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
5 1 2 3 4 5 2 ADD 2 4 1 MIN 4 5
Sample Output
5
/*
poj3580 splay树 REVOVLE循环
给定一个数列:a1,a2,.... an
进行以下6种操作：
ADD x y D : 给第x个数到第y个数加D
REVERSE x y : 反转[x,y]
REVOVLE x y T : 对[x,y]区间的数循环右移T次 (这个最开始没想到这么弄)
(先把T对长度取模，然后相当于把[y-T+1,y]放到[x,y-T] 的前面，T可能出现负数或者特别大)
INSERT x P : 在第x个数后面插入P
DELETE x : 删除第x个数
MIN x y : 查询[x,y]之间的最小的数
像min，add是参照rev，size来写。然后就只有第三个操作可以一开始想不到，其它大致就是各种基本操作的组合了
最开始写错del导致TLE几次
hhh-2016-02-21 03:06:02
*/
#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;
#define key_value ch[ch[root][1]][0]
const int maxn = 200010;
int ch[maxn][2];
int pre[maxn],siz[maxn],num[maxn];
int rev[maxn],key[maxn];
int add[maxn];
int Min[maxn],a[maxn];
int tot;
int root;
void push_up(int r)
{
int lson = ch[r][0],rson = ch[r][1];
siz[r] = siz[lson] + siz[rson] + 1;
Min[r] = min(key[r],min(Min[lson],Min[rson]));
}
void update_add(int r,int val)
{
if(!r) return;
key[r] += val;
add[r] += val;
Min[r] += val;
}
void update_rev(int r)
{
if(!r)return ;
swap(ch[r][0],ch[r][1]);
rev[r] ^= 1;
}
void push_down(int r)
{
if(rev[r])
{
update_rev(ch[r][0]);
update_rev(ch[r][1]);
rev[r] = 0;
}
if(add[r])
{
update_add(ch[r][0],add[r]);
update_add(ch[r][1],add[r]);
add[r] = 0;
}
}
void NewNode(int &r,int far,int k)
{
r = ++tot; //不能为0
pre[r] = far;
ch[r][0] = ch[r][1] = 0;
siz[r] = 1;
Min[r] = k;
key[r] = k;
rev[r] = 0;
add[r] = 0;
}
void rotat(int x,int kind)
{
int y = pre[x];
push_down(y);
push_down(x);
ch[y][!kind] = ch[x][kind];
pre[ch[x][kind]] = y;
if(pre[y])
ch[pre[y]][ch[pre[y]][1]==y] = x;
pre[x] = pre[y];
ch[x][kind] = y;
pre[y] = x;
push_up(y);
}
void build(int &x,int l,int r,int far)
{
if(l > r) return ;
int mid = (l+r) >>1;
NewNode(x,far,a[mid]);
build(ch[x][0],l,mid-1,x);
build(ch[x][1],mid+1,r,x);
push_up(x);
}
void splay(int r,int goal)
{
push_down(r);
while(pre[r] != goal)
{
if(pre[pre[r]] == goal)
{
push_down(pre[r]);
push_down(r);
rotat(r,ch[pre[r]][0] == r);
}
else
{
push_down(pre[pre[r]]);
push_down(pre[r]);
push_down(r);
int y = pre[r];
int kind = ch[pre[y]][0] == y;
if(ch[y][kind] == r)
{
rotat(r,!kind);
rotat(r,kind);
}
else
{
rotat(y,kind);
rotat(r,kind);
}
}
}
push_up(r);
if(goal == 0)
root = r;
}
int get_kth(int r,int k)
{
push_down(r);
int t = siz[ch[r][0]] + 1;
if(k == t)return r;
if(t > k) return get_kth(ch[r][0],k);
else return get_kth(ch[r][1],k-t);
}
int get_next(int r)
{
push_down(r);
if(ch[r][1] == 0)return -1;
r = ch[r][1];
while(ch[r][0])
{
r = ch[r][0];
push_down(r);
}
return r;
}
void Reverse(int l,int r)
{
splay(get_kth(root,l),0);
splay(get_kth(root,r+2),root);
update_rev(key_value);
push_up(ch[root][1]);
push_up(root);
}
void Add(int l,int r,int val)
{
splay(get_kth(root,l),0);
splay(get_kth(root,r+2),root);
update_add(key_value,val);
push_up(ch[root][1]);
push_up(root);
}
void ini(int n)
{
tot = root = 0;
ch[root][0] = ch[root][1] = pre[root] = siz[root] = num[root] = 0;
Min[root] = 0x3f3f3f3f;
rev[root] = add[root] = 0;
NewNode(root,0,-1);
NewNode(ch[root][1],root,-1);
for(int i=1; i <= n; i++)
{
scanf("%d",&a[i]);
}
build(key_value,1,n,ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
int get_min(int r)
{
push_down(r);
while(ch[r][0])
{
r = ch[r][0];
push_down(r);
}
return r;
}
void Delete(int r)
{
splay(get_kth(root,r+1),0);
if(ch[root][0] == 0 || ch[root][1] == 0)
{
root = ch[root][0] + ch[root][1];
pre[root] = 0;
return;
}
int k = get_min(ch[root][1]);
splay(k,root);
ch[ch[root][1]][0] = ch[root][0];
root = ch[root][1];
pre[ch[root][0]] = root;
pre[root] = 0;
push_up(root);
}
void Insert(int x,int y)
{
splay(get_kth(root,x+1),0);
splay(get_kth(root,x+2),root);
NewNode(key_value,ch[root][1],y);
push_up(ch[root][1]);
push_up(root);
}
int MIN(int x,int y)
{
splay(get_kth(root,x),0);
splay(get_kth(root,y+2),root);
push_up(ch[root][1]);
push_up(root);
return Min[key_value];
}
void Revovle(int x,int y,int T)
{
splay(get_kth(root,y-T+1),0);
splay(get_kth(root,y+2),root);
int tmp = key_value;
//pre[key_value] = 0;
key_value = 0;
push_up(ch[root][1]);
push_up(root);
splay(get_kth(root,x),0);
splay(get_kth(root,x+1),root);
key_value = tmp;
pre[key_value] = ch[root][1];
push_up(ch[root][1]);
push_up(root);
}
int main()
{
int n,p;
while(scanf("%d",&n) != EOF)
{
ini(n);
scanf("%d",&p);
char opr[20];
int x,y,z;
while(p--)
{
scanf("%s",opr);
if(strcmp(opr,"ADD") == 0)
{
scanf("%d%d%d",&x,&y,&z);
Add(x,y,z);
}
else if(strcmp(opr,"INSERT") == 0)
{
scanf("%d%d",&x,&y);
Insert(x,y);
}
else if(strcmp(opr,"DELETE") == 0)
{
scanf("%d",&x);
Delete(x);
}
else if(strcmp(opr,"MIN") == 0)
{
scanf("%d%d",&x,&y);
printf("%d
",MIN(x,y));
// for(int i =1;i <= 5;i++)
// printf("%d
",Min[i]);
}
else if(strcmp(opr,"REVERSE") == 0)
{
scanf("%d%d",&x,&y);
Reverse(x,y);
}
else if(strcmp(opr,"REVOLVE") == 0)
{
scanf("%d%d%d",&x,&y,&z);
int t = (y-x+1);
z = (z%t+t)%t;
Revovle(x,y,z);
}
}
}
return 0;
}