Looploop
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1651 Accepted Submission(s): 517
Problem Description
XXX gets a new toy named Looploop. The toy has N elements arranged in a loop, an arrow pointing to one of the elements, and two preset parameters k1 and k2. Every element has a number on it.
The figure above shows a Looploop of 6 elments. Let's assuming the preset parameter k1 is 3, and k2 is 4.
XXX can do six operations with the toy.
1: add x
Starting from the arrow pointed element, add x to the number on the clockwise first k2 elements.
2: reverse
Starting from the arrow pointed element, reverse the first k1 clockwise elements.
3: insert x
Insert a new element with number x to the right (along clockwise) of the arrow pointed element.
4: delete
Delete the element the arrow pointed and then move the arrow to the right element.
5: move x
x can only be 1 or 2. If x = 1 , move the arrow to the left(along the counterclockwise) element, if x = 2 move the arrow to the right element.
6: query
Output the number on the arrow pointed element in one line.
XXX wants to give answers to every query in a serial of operations.
Input
There are multiple test cases.
For each test case the first line contains N,M,k1,k2(2≤k1<k2≤N≤105, M≤105) indicating the initial number of elements, the total number of operations XXX will do and the two preset parameters of the toy.
Second line contains N integers ai(-104≤ai≤104) representing the N numbers on the elements in Looploop along clockwise direction. The arrow points to first element in input at the beginning.
Then m lines follow, each line contains one of the six operations described above.
It is guaranteed that the "x" in the "add","insert" and "move" operations is always integer and its absolute value ≤104. The number of elements will never be less than N during the operations.
The input ends with a line of 0 0 0 0.
For each test case the first line contains N,M,k1,k2(2≤k1<k2≤N≤105, M≤105) indicating the initial number of elements, the total number of operations XXX will do and the two preset parameters of the toy.
Second line contains N integers ai(-104≤ai≤104) representing the N numbers on the elements in Looploop along clockwise direction. The arrow points to first element in input at the beginning.
Then m lines follow, each line contains one of the six operations described above.
It is guaranteed that the "x" in the "add","insert" and "move" operations is always integer and its absolute value ≤104. The number of elements will never be less than N during the operations.
The input ends with a line of 0 0 0 0.
Output
For each test case, output case number in the first line(formatted as the sample output). Then for each query in the case, output the number on the arrow pointed element in a single line.
Sample Input
5 1 2 4
3 4 5 6 7
query
5 13 2 4
1 2 3 4 5
move 2
query
insert 8
reverse
query
add 2
query
move 1
query
move 1
query
delete
query
0 0 0 0
3 4 5 6 7
query
5 13 2 4
1 2 3 4 5
move 2
query
insert 8
reverse
query
add 2
query
move 1
query
move 1
query
delete
query
0 0 0 0
Sample Output
Case #1:
3
Case #2:
2
8
10
1
5
1
3
Case #2:
2
8
10
1
5
1
/*
hdu 4453 splay
把环换成链然后进行splay即可
主要是move那个操作开始想复杂了，只需要直接进行删除插入
add、reverse、insert、delete、query
move 1：把第n+1个点删除，然后放入第一个点之后。
move 2：把第2个点删除，然后放入第n+1个点后。
hhh-2016-02-21 07:01:08
*/ #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,tp;
int root,n;
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 inOrder(int r)
{
if(!r)
return;
inOrder(ch[r][0]);
printf("%d ",key[r]);
inOrder(ch[r][1]);
} void debug()
{
inOrder(root);
cout <<endl;
} 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;
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;
} int Delete(int r)
{
int t = get_kth(root,r+1);
splay(t,0);
if(ch[root][0] == 0 || ch[root][1] == 0)
{
root = ch[root][0] + ch[root][1];
pre[root] = 0;
return key[t];
}
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);
n--;
return key[t];
} void Insert(int x,int y)
{
splay(get_kth(root,x),0);
splay(get_kth(root,x+1),root);
NewNode(key_value,ch[root][1],y);
push_up(ch[root][1]);
push_up(root);
n++;
} void Move(int x)
{
if(x == 1)
{
int t = Delete(n);
Insert(1,t);
// debug();
}
else
{
int t = Delete(1);
Insert(n+1,t);
}
} int main()
{
int p,k1,k2;
int cas = 1;
while(scanf("%d%d%d%d",&n,&p,&k1,&k2) != EOF)
{
if(!n && !p && !k1 && !k2)
break;
printf("Case #%d:\n",cas++);
ini(n);
char opr[10];
int x;
for(int i =1; i <= p; i++)
{
scanf("%s",opr);
if(opr[0] == 'a')
{
scanf("%d",&x);
Add(1,k2,x);
}
else if(opr[0] == 'm')
{
scanf("%d",&x);
Move(x);
}
else if(opr[0] == 'r')
{
Reverse(1,k1);
}
else if(opr[0] == 'q')
{
printf("%d\n",key[get_kth(root,2)]);
}
else if(opr[0] == 'i')
{
scanf("%d",&x);
Insert(2,x);
}
else if(opr[0] == 'd')
{
Delete(1);
}
}
}
return 0;
}