POJ 2828 Buy Tickets（线段树 ）

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 nextN lines contain the pairs of values Pos_i and Val_i in the increasing order ofi (1 ≤ i ≤ N). For each i, the ranges and meanings ofPos_i and Val_i are as follows:

• Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind thePos_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 valueVal_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.

题意：n个操作，每次在第k个位置插入一个数，最后输出排序结果

分析：对操作从后往前推，因为后面的操作会影响当先位置，所以求出后面的操作位置，从而推出当先序列剩下位置中第k位，从而插入进去，然后更新。

线段树中保存区间有多少位置已经插入，和k比较。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <queue>#define mem(p,k) memset(p,k,sizeof(p));#define rep(i,j,k) for(int i=j; i<k; i++)#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define inf 0x6fffffff#define ll long longusing namespace std;const int mod=1e9+7;int a[210000],b[210000];int n,m,k,maxx,cur,x,y;int minn;int nex[4][2]={0,1,1,0,0,-1,-1,0};int sum[210000<<2],num[210000<<2];void query(int k,int c,int l,int r,int rt){    if(l==r){        num[rt]=c;        sum[rt]=1;        return;    }    int m=(l+r)>>1,f=m-l+1-sum[rt<<1];    if(k<=f){        query(k,c,lson);    }    else query(k-f,c,rson);    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void print(int l,int r,int rt){    if(l==r){        cout<<num[rt];        if(++cur<n)cout<<" ";        else cout<<endl;        return;    }    int m=(l+r)>>1;    print(lson);    print(rson);}int main(){    while(cin>>n){            rep(i,0,n){               scanf("%d%d",a+i,b+i);            }            mem(sum,0);            for(int i=n-1;i>=0;i--){                query(a[i]+1,b[i],1,n,1);            }            cur=0;            print(1,n,1);    }    return 0;}//freopen("C:\\Users\\LENOVO\\Desktop\\read.txt","r",stdin);