hdu 3436 线段树 一顿操作
来源:互联网 发布:淘宝流量高峰时段 编辑:程序博客网 时间:2024/05/01 17:04
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.
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
39 5Top 1Rank 3Top 7Rank 6Rank 86 2Top 4Top 57 4Top 5Top 2Query 1Rank 6
Sample Output
Case 1:358Case 2:Case 3:36
/*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;}
0 0
- hdu 3436 线段树 一顿操作
- hdu-4578(线段树操作)
- HDU 3954 线段树 特殊LAZY操作
- HDU 4578 线段树 多lazy操作
- hdu 4578 线段树多重操作
- hdu 3333 线段树离线操作
- hdu 4630 线段树+在线操作
- hdu 2871 线段树(各种操作)
- hdu 3577 线段树区间操作(带有懒操作)
- hdu 3397 Sequence operation(线段树的各种操作)
- hdu 1698 Just a Hook (线段树区间操作,)
- hdu 3954(线段树的特殊lazy操作)
- hdu-4587-线段树的区间操作- lazy标记
- HDU 3911 线段树 LAZY操作+成段更新
- HDU 2795 Billboard (RMQ线段树&合并操作技巧)
- hdu 4358 Boring counting 线段树离线操作
- hdu-4605-Magic Ball Game-线段树+离线操作
- hdu 4630 No Pain No Game(线段树+离线操作)
- 欢迎使用CSDN-markdown编辑器
- java中list集合的clear
- CentOS6.5虚拟机下搭建SVN服务器
- Ubuntu基本命令整理
- c++类构造函数、析构函数与虚函数之间的那点小事
- hdu 3436 线段树 一顿操作
- 循环链表
- RecyclerView
- UIImage imageName方法扩充提醒功能
- UI更新方法Handler和runOnUiThread
- 10046事件与tkprof命令
- service
- C++问题小结--3.构造函数与析构函数关于其自身特点的简单代码陈述
- shape