POJ 2828 Buy Tickets（线段树）

原题链接

Problem Description

Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…

The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.

It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!

People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.

Input

There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:

Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
There no blank lines between test cases. Proceed to the end of input.

Output

For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.

Sample Input

4
0 77
1 51
1 33
2 69
4
0 20523
1 19243
1 3890
0 31492

Sample Output

77 33 69 51
31492 20523 3890 19243

题目大意

每次操作给出a b，代表将编号为b的人插入到a位置的后部。最后要求输出生成的队列。

解题思路

本来是想练Splay的结果发现这就是个线段树的单点修改???每个结点只需要维护size即可，在pushdown的过程中也是根据size来决定是向左儿子走还是向右儿子走。

AC代码

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<algorithm>#include<cmath>#include<vector>#include<string>#include<queue>#include<list>#include<stack>#include<set>#include<map>#define ll long long#define ull unsigned long long#define rep(i,n) for(int i = 0;i < n; i++)#define fil(a,b) memset((a),(b),sizeof(a))#define cl(a) fil(a,0)#define pb push_back#define mp make_pair#define exp 2.7182818#define PI 3.141592653589793//#define inf 0x3f3f3f3f#define fi first#define se second#define eps 1e-8#define mod 2000000014ll#define sign(x) ((x)>eps?1:((x)<-eps?(-1):(0)))using namespace std;double mysqrt(double x) { return max(0.0, sqrt(x)); }const int maxnode=4e5+10;int cap[600010];int out[200010];int in[200010][2];int pos,val;void build(int o,int l,int r){    if(l==r)    {        cap[o]=1;        return ;    }    int lson=o<<1;    int rson=o<<1|1;    int mid=(l+r)/2;    cap[o]=r-l+1;    build(lson,l,mid);    build(rson,mid+1,r);}void insert(int o,int l,int r){    //cout<<pos<<"zz"<<endl;    if(l==r)    {        cap[o]=0;        out[l]=val;        return ;    }    int lson=o<<1;    int rson=o<<1|1;    int mid=(l+r)/2;    if(pos<=cap[lson])    {        insert(lson,l,mid);    }    else    {        pos-=cap[lson];        insert(rson,mid+1,r);    }    cap[o]--;}int main() {      int n;    //int pos,val;    while(cin>>n)    {        build(1,1,n);        for(int i=1;i<=n;++i)        {            scanf("%d%d",&in[i][0],&in[i][1]);        }        for(int i=n;i>=1;--i)        {            pos=in[i][0]+1;            val=in[i][1];            //cout<<pos<<"hah"<<endl;            insert(1,1,n);        }        for(int i=1;i<=n;++i) printf("%d ",out[i]);        cout<<endl;    }    return 0;  }