POJ 2828 Buy Tickets

http://poj.org/problem?id=2828

Buy Tickets

Time Limit: 4000MSMemory Limit: 65536KTotal Submissions: 14423Accepted: 7210

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 Pos_i and Val_i in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

• Pos_i ∈ [0, i ? 1] — The i-th person came to the queue and stood right behind the Pos_i-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.
• Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

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

40 771 511 332 6940 205231 192431 38900 31492

Sample Output

77 33 69 5131492 20523 3890 19243

Hint

The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.

Source

POJ Monthly--2006.05.28, Zhu, Zeyuan

题目大意就是添加数据 不断插入数据 每个数据两个值 pos 位置（代表要插入的位置），val 价值。

最后输出插入多组数据后的序列（val序列）

可以用线段树 叶子结点存最终的序列 每个节点增加个pos域，代表此区域有pos个位置可以插入。然后逆序进行更新。

以样例为例

 

红色的数字代表可插入的位置个数。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn  200100  //这是为什么 改为200100+5就re
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;
struct node
{
int l,r;
int pos,val;
}nod[maxn*3];
int pos[maxn];
int k;
struct pair
{
int pos,val;
}mp[maxn];
void build(int i,int l,int r)
{
nod[i].l=l;
nod[i].r=r;
if(l==r)
{
nod[i].pos=1;
return ;
}
int mid=(l+r)/2;
build(i<<1,l,mid);
build(i<<1|1,mid+1,r);
nod[i].pos=nod[i<<1].pos+nod[i<<1|1].pos;
}

void update(int i,int pos,int val)
{
if(nod[i].l==nod[i].r)
{
nod[i].val=val;
nod[i].pos=0;
return ;
}
if(nod[i<<1].pos>pos)
{
update(i<<1,pos,val);
}
else
{
update(i<<1|1,pos-nod[i<<1].pos,val);//!!
}
nod[i].pos=nod[i<<1].pos+nod[i<<1|1].pos;
}

void query(int i)
{
if(nod[i].l==nod[i].r)
{
pos[k++]=nod[i].val;
return ;
}
query(i<<1);
query(i<<1|1);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(~scanf("%d",&n))
{
k=0;
for(int i=0;i<n;i++)
scanf("%d %d",&mp[i].pos,&mp[i].val);//first 是pos second 是val
build(1,1,n);
for(int i=n-1;i>=0;i--)
{
update(1,mp[i].pos,mp[i].val);
}
query(1);
for(int i=0;i<n-1;i++)
printf("%d ",pos[i]);
printf("%d\n",pos[n-1]);
}
    return 0;
}