hdu 3436 线段树 一顿操作

Queue-jumpers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3348 Accepted Submission(s): 904



Problem Description

Ponyo and Garfield are waiting outside the box-office for their favorite movie. Because queuing is so boring, that they want to play a game to kill the time. The game is called “Queue-jumpers”. Suppose that there are N people numbered from 1 to N stand in a
line initially. Each time you should simulate one of the following operations:

1. Top x :Take person x to the front of the queue

2. Query x: calculate the current position of person x

3. Rank x: calculate the current person at position x

Where x is in [1, N].

Ponyo is so clever that she plays the game very well while Garfield has no idea. Garfield is now turning to you for help.

Input

In the first line there is an integer T, indicates the number of test cases.(T<=50)

In each case, the first line contains two integers N(1<=N<=10^8), Q(1<=Q<=10^5). Then there are Q lines, each line contain an operation as said above.

Output

For each test case, output “Case d:“ at first line where d is the case number counted from one, then for each “Query x” operation ,output the current position of person x at a line, for each “Rank x” operation, output the current person at position x at a line.

Sample Input

3
9 5
Top 1
Rank 3
Top 7
Rank 6
Rank 8
6 2
Top 4
Top 5
7 4
Top 5
Top 2
Query 1
Rank 6

Sample Output

Case 1:
3
5
8
Case 2:
Case 3:
3
6

/*
hdu 3436 线段树 一顿操作

这个题以前用splay树做过,但是最近练习线段树中(据说线段树能解决splay树中的很多操作)
Top:     将第x个数移动到队首
Query:   查询x的位置
Rank:    找出排第x的数

top想的数之间在线段树前面预留一部分,于是线段树最多需要2e5.对于Rank可以直接查找
主要是对于Query没想到什么办法.最后直接是用数组来保存每个数的位置,然后利用求和便能得到
某数在这个队列中的位置

hhh-2016-04-12 19:20:16
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <functional>
typedef long long ll;
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;

const int maxn = 1e6+10;

struct node
{
int l,r;
int sum,val;
int mid()
{
return (l+r)>>1;
}
} tree[maxn<<2];
int T,n,m;
int a[maxn];
int ano[maxn];
int st[maxn],en[maxn];
int pos[maxn];
char op[maxn][6];
int tot,TOT;

void push_up(int i)
{
tree[i].sum = tree[lson].sum + tree[rson].sum;
}

void build(int i ,int l,int r)
{
tree[i].l =l ,tree[i].r = r;
tree[i].sum = 0;
tree[i].val = 0;
if(l == r)
{
if(tree[i].l > m)
{
int t = tree[i].l-m;
tree[i].val = t;
tree[i].sum = en[t]-st[t]+1;
}
return ;
}
int mid = tree[i].mid();
build(lson,l,mid);
build(rson,mid+1,r);
push_up(i);
}

void push_down(int i)
{

}

void update(int i,int k,int val)
{
if(tree[i].l == tree[i].r )
{
if(!val) tree[i].sum = 0,tree[i].val = 0;
else tree[i].sum = en[val]-st[val]+1,tree[i].val = val;
return ;
}
int mid = tree[i].mid();
if(k <= mid)
update(lson,k,val);
else
update(rson,k,val);
push_up(i);
}

int sum(int i,int l,int r)
{
if(tree[i].l>=l && tree[i].r <= r)
return tree[i].sum;
int mid = tree[i].mid();
int su = 0;
if(l <= mid)
su += sum(lson,l,r);
if(r > mid)
su += sum(rson,l,r);
push_up(i);
return su;
}

int get_k(int i,int k)
{
if(tree[i].l == tree[i].r && k <= en[tree[i].val]-st[tree[i].val]+1)
return st[tree[i].val]+k-1;
int mid = tree[i].mid();

if(k <= tree[lson].sum)
return get_k(lson,k);
else
return get_k(rson,k-tree[lson].sum);
push_up(i);
}

int bin(int key)
{
int l = 1,r = tot-1;
while(l <= r)
{
int mid = (l+r)>>1;
if(st[mid]<=key && en[mid]>=key)
return mid;
else if(key < st[mid])
r = mid - 1;
else
l = mid + 1;
}
}

int main()
{
int cas = 1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
tot = 0;
for(int i = 1; i <= m; i++)
{
scanf("%s%d",op[i],&a[i]);
if(op[i][0] == 'T' || op[i][0] == 'Q')
ano[tot++] = a[i];
}
ano[tot++] = 1,ano[tot++] = n;
sort(ano,ano+tot);
//        for(int i = 0;i < tot;i++)
//            cout << ano[i] <<" ";
//        cout << endl;
TOT = tot;
tot = 1;
st[tot] = ano[0],en[tot] = ano[0];
tot++;
printf("Case %d:\n",cas++);
for(int i = 1; i < TOT; i++)
{
if(ano[i] != ano[i-1])
{
if(ano[i] - ano[i-1] > 1)
{
st[tot] = ano[i-1]+1;
en[tot++] = ano[i]-1;
}
st[tot] = ano[i];
en[tot] = ano[i];
tot++;
}
}
//        for(int i = 1;i < tot;i++)
//            cout <<st[i] << " "<<en[i] <<endl;
build(1,1,m+tot-1);
memset(ano,0,sizeof(ano));
int cur = m;
for(int i = 1; i <= m; i++)
{
int tp = bin(a[i]);
//cout << "val:" << a[i] << " "<<tp<<endl;
if(op[i][0] == 'T')
{
if(ano[tp])
{
update(1,pos[tp],0);
update(1,cur,tp);
pos[tp] = cur;
cur --;
}
else
{
update(1,m+tp,0);
update(1,cur,tp);
pos[tp] = cur;
cur--;
ano[tp] = 1;
}
}
else if(op[i][0] == 'Q')
{
if(ano[tp])
{
printf("%d\n",sum(1,1,pos[tp]));
}
else
{
printf("%d\n",sum(1,1,m+tp));
}
}
else
{
printf("%d\n",get_k(1,a[i]));
}
}
}
return 0;
}