POJ 2828Buy Tickets(线段树的单点维护)

Posted 2020-09-29 欲儿很轻狂

Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 20462		Accepted: 10096

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

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

解题思路：

一个不错的排队问题，可通过从后面开始排队来进行巧妙地排队解决，假设后面的人都已经站在正确的位置上了，那么到那个人站的时候，现在的位置上已经都是后面的那些人了，只要数pos个空格，那那个人站的位置能确定了。确定之后就可以求下一个了，所以这个前提和结论都成立了。所以我们只要从后面人站起，数pos个空格站上去就行了。 具体实现看代码吧.

PS:

G++  TLE, C++ AC.

 

AC代码：

#include<cstring>
#include<string>
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
const int maxsize = 200002;
int cnt,n,sum[maxsize*4],ans[maxsize*4];
struct node
{
    int l,r;
} tree[maxsize*4];
void build(int l,int r,int root)
{
    if(l==r)
    {
        sum[root]=1;
        return ;
    }
    int mid=(l+r)/2;
    build(lson);
    build(rson);
    sum[root]=sum[root*2]+sum[root*2+1];
}
void Update(int pos,int b,int l,int r,int root)
{
    if(l==r)
    {
        if(sum[root])
        {
            sum[root]=0;
            ans[root]=b;
        }
        return ;
    }
    int mid=(l+r)/2;
    if(pos<=sum[root*2])  Update(pos,b,lson);  //前面的话就霸占
    else    Update(pos-sum[root*2],b,rson); //后面的话就往后滚
    sum[root]=sum[root*2]+sum[root*2+1];
}
void Printf(int l,int r,int root)
{
    if(l==r)
    {
        cnt++;
        printf("%d%c",ans[root],cnt==n?\'\\n\':\' \');
        return ;
    }
    int mid=(l+r)/2;
    Printf(lson);
    Printf(rson);
}
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n&&n)  //(cin>>n,n)就TLE，无语
    {
        cnt=0;
        build(1,n,1);
        for(int i=1; i<=n; i++)
        {
            scanf("%d %d",&tree[i].l,&tree[i].r);
            tree[i].l++;
        }
        ///pos记得加1
        for(int i=n; i>=1; i--)   Update(tree[i].l,tree[i].r,1,n,1);
        Printf(1,n,1);
    }
    return 0;
}