首页 > 代码库 > poj2828 Buy ticket

poj2828 Buy ticket

2024-07-12 18:00:23   220人阅读

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

40 771 511 332 6940 205231 192431 38900 31492

Sample Output

77 33 69 5131492 20523 3890 19243
题目大意 ：给出n个人的排队信息，ai和bi，ai表示第i个人要排在当前第ai个人的后面，bi为第i个人的标识，求出最后的队伍排列。
思路：先把每个人的信息都记录下来，再反着插入即可。用线段树记录当前区间还有多少空位。
 1 /* 2  * Author:  Joshua 3  * Created Time:  2014年07月15日 星期二 15时51分53秒 4  * File Name: poj2828.cpp 5  */ 6 #include<cstdio> 7 #include<cstring> 8 #define maxn 200020 9 #define L(x) (x<<1)10 #define R(x) (x<<1 |1)11 struct node12 {13     int l,r,cnt;14 } e[maxn<<2];15 int n;16 int pos[maxn],val[maxn],ans[maxn]; 17 void built(int t,int l,int r)18 {19     e[t].l=l;20     e[t].r=r;21     e[t].cnt=r-l+1;22     if (l==r) return;23     int mid=l+((r-l+1)>>1)-1;24     built(L(t),l,mid);25     built(R(t),mid+1,r);26 }27 28 void updata(int t,int x,int v)29 {30     --e[t].cnt;31     if (e[t].l==e[t].r)32     {33         ans[e[t].l]=v;34         return;35     }36     if (e[L(t)].cnt>x)37         updata(L(t),x,v);38     else39         updata(R(t),x-e[L(t)].cnt,v);40 }41 void solve()42 {43     for (int i=1;i<=n;++i)44        scanf("%d%d",&pos[i],&val[i]);45     built(1,1,n);46     for (int i=n;i>0;--i)47        updata(1,pos[i],val[i]);48     for (int i=1;i<=n;++i)49        printf("%d%c",ans[i],i==n ? ‘\n‘:‘ ‘);        50 }51 52 int main()53 {54     while (scanf("%d",&n)==1)55         solve();56     return 0;57 }

 


联系
我们