hdu 3436 splay树+离散化*
Queue-jumpers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3139 Accepted Submission(s): 848
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.
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.
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
Author
wzc1989
Source
/*
hdu 3436 splay树+离散化*
本来以为很好做的，写到中途发现10^8,GG
然后参考了下，把操作不用的区间缩点离散化处理
然后就是删除点，感觉自己开始写的太麻烦了，将要删除的点移动到根，如果没有儿子直接删掉，
否则将右树的最小点移到ch[r][1]使右树没有左子树，然后把根的左树接到右树上
hhh-2016-02-20 22:22:22
*/
#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],key[maxn],siz[maxn],num[maxn];
int root,tot,cnt,n,TOT;
int posi[maxn];
char qry[maxn][10];
int op[maxn];
int te[maxn];
int s[maxn],e[maxn];
void Treaval(int x) {
if(x) {
Treaval(ch[x][0]);
printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,key = %2d num= %2d \n",x,ch[x][0],ch[x][1],pre[x],siz[x],key[x],num[x]);
Treaval(ch[x][1]);
}
}
void debug() {printf("%d\n",root);Treaval(root);}
void push_up(int r)
{
int lson = ch[r][0],rson = ch[r][1];
siz[r] = siz[lson] + siz[rson] + num[r];
}
void push_down(int r)
{
}
void inOrder(int r)
{
if(!r)return;
inOrder(ch[r][0]);
printf("%d ",key[r]);
inOrder(ch[r][1]);
}
void NewNode(int &r,int far,int k)
{
r = ++tot;
posi[k] = r;
key[r] = k;
pre[r] = far;
ch[r][0] = ch[r][1] = 0;
siz[r] = num[r] = e[k]-s[k]+1;
}
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,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 Bin(int x)
{
int l = 0,r = TOT-1;
while(l<=r)
{
int mid=(l+r)>>1;
if(s[mid]<=x&&e[mid]>=x)
return mid;
if(e[mid]<x)
l=mid+1;
else
r=mid-1;
}
}
int get_min(int r)
{
push_down(r);
while(ch[r][0])
{
r = ch[r][0];
push_down(r);
}
return r;
}
int get_kth(int r,int k)
{
int t = siz[ch[r][0]];
if(k<=t)
return get_kth(ch[r][0],k);
else if(k<=t+num[r])
return s[key[r]]+(k-t)-1;
else
return get_kth(ch[r][1],k-t-num[r]);
}
void delet()
{
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);
}
int top(int t)
{
int r = Bin(t);
r = posi[r];
splay(r,0);
delet();
splay(get_min(root),0);
ch[r][0] = 0;
ch[r][1] = root;
pre[root] = r;
root = r;
pre[root] = 0;
push_up(root);
// debug();
}
int Query(int x)
{
int r = Bin(x);
r = posi[r];
splay(r,0);
return siz[ch[r][0]]+1;
}
int get_rank(int x,int k)
{
int t = siz[ch[x][0]];
if(k <= t)
return get_rank(ch[x][0],k);
else
return get_rank(ch[x][1],k-t);
}
void ini(int n)
{
tot = root = 0;
ch[root][0] = ch[root][1] = pre[root] = siz[root] = num[root] = 0 ;
build(root,0,n-1,0);
push_up(ch[root][1]);
push_up(root);
//inOrder(root);
}
int main()
{
int q,T;
int cas =1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&q) ;
if(n == -1 && q == -1)
break;
int tcn = 0;
printf("Case %d:\n",cas++);
for(int i =1; i <= q; i++)
{
scanf("%s%d",qry[i],&op[i]);
if(qry[i][0] == 'T' || qry[i][0] == 'Q')
te[tcn++] = op[i];
}
te[tcn++] = n;
te[tcn++] = 1;
sort(te,te+tcn);
TOT= 0;
s[TOT] = te[0],e[TOT] = te[0],TOT++;
for(int i = 1; i < tcn; i++)
{
if(te[i] != te[i-1] && i)
{
if(te[i] - te[i-1] > 1)
{
s[TOT] = te[i-1]+1;
e[TOT] = te[i]-1;
TOT++;
}
s[TOT] = te[i];
e[TOT] = te[i];
TOT++;
}
}
ini(TOT);
//debug();
for(int i = 1; i <= q; i++)
{
if(qry[i][0]=='T')
top(op[i]);
else if(qry[i][0]=='Q')
printf("%d\n", Query(op[i]));
else
printf("%d\n",get_kth(root,op[i]));
}
//debug();
}
return 0;
}
