POJ 2828 Buy Tickets (线段树，逆序遍历)

Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 13017		Accepted: 6449

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

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

 

题目的意思就是给定N个插队的顺序，求最后队伍的排序

看了下题目给的数据20W，链表（寻址）和数组（数组移动）会超时。

看了题解才知道用线段树。线段树可以精确查找到位置。

读入插队顺序的时候如果从前往后读的话，每次插入后，有的人的当前位置会改变。

所以采用从后往前读的方法

因为因为最后一个人的位置是固定的，所以从后往前排，每次排的那个人的位置就固定了。

拿样例分析一下：

第一次在第0位插：77

第二次在第一位插51：77 51

第三次在第一位插33：77 33 51

第四次在第二位插69：77 33 69 51

 

逆序的话：

第一次在前面还有2个空位的地方插69：_ _ 69 _

第二次在前面还有1个空位的地方插33：_ 33 69 _

第三次在前面还有1个空位的地方插51：_ 33 69 51

第四次在前面还有0个空位的地方插77:77 33 69 51

不过这道题目的想法很重要，每个节点表示这个区间内的空的位置的数量，从后往前读，pos的值表示这个人前面有多少个空位，然后在树里面查找空位。

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=200000+5;
struct node
{
    int pos;
    int val;
    int l,r;
    int mid()
    {
        return (l+r)/2;
    }
}a[MAXN*4];

struct input
{
    int pos;
    int val;
}num[MAXN];

int ans[MAXN],cnt=1;

void btree(int l,int r,int step)
{
    a[step].l=l;
    a[step].r=r;
    if(l==r)
    {
        a[step].pos=1;
        return ;
    }
    int mid=a[step].mid();
    btree(l,mid,step*2);
    btree(mid+1,r,step*2+1);
    a[step].pos=a[step*2].pos+a[step*2+1].pos;
}

void ptree(int step,int pos,int val)
{
    if(a[step].l==a[step].r)
    {
        a[step].val=val;
        a[step].pos=0;
        return ;
    }
    if(a[step*2].pos>pos)
        ptree(step*2,pos,val);
    else
        ptree(step*2+1,pos-a[step*2].pos,val);
    a[step].pos=a[step*2].pos+a[step*2+1].pos;
}

void fintree(int step)
{
    if(a[step].l==a[step].r)
    {
        ans[cnt++]=a[step].val;
        return ;
    }
    fintree(step*2);
    fintree(step*2+1);
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d %d",&num[i].pos,&num[i].val);
        btree(1,n,1);
        for(int i=n;i>=1;i--)
            ptree(1,num[i].pos,num[i].val);
        cnt=1;
        fintree(1);
        for(int i=1;i<n;i++)
            printf("%d ",ans[i]);
        printf("%d
",ans[n]);
    }
    return 0;
}

代码跑起来有三千七百多秒，还要再优化一下。