本文主要是介绍POJ 2828 Buy Tickets (线段树，单点更新)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
http://poj.org/problem?id=2828
| Buy Tickets
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 Valiin the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:
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 |
题意
告诉你每个人进队时插在第几个人后面，问最终的队列结果。
分析：
最后一个进队的位置才能即刻确定下来，我们要从后往前操作。比如这个人插在当前第x个人的后面，那就意味着他前面要留出x个空位（反向操作，前面的人还没进队）。我们用线段树维护当前队列区间内剩余的空位，单点更新。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn 200007using namespace std;itn pos[maxn],val[maxn];
int ans[maxn];
int rest[maxn<<2];void build(int k,int l,int r)
{rest[k]=r-l;if (r-l==1) return ;build(k*2+1,l,l+r>>1);build(k*2+2,l+r>>1,r);
}int query(int _rest,int k,int l,int r)
{rest[k]--;if (r-l==1) return l;if (_rest<rest[k*2+1])//左儿子区间内可放{return query(_rest,k*2+1,l,l+r>>1);}else//左儿子放不下，放在右儿子里{return query(_rest-rest[k*2+1],k*2+2,l+r>>1,r);}
}int main()
{#ifndef ONLINE_JUDGEfreopen("/home/fcbruce/文档/code/t","r",stdin);#endif // ONLINE_JUDGEint n;while (~scanf("%d",&n)){for (itn i=0;i<n;i++){scanf("%d %d",pos+i,val+i);}build(0,0,n);for (int i=n-1;i>=0;i--){ans[query(pos[i],0,0,n)]=val[i];}for (int i=0;i<n;i++)printf(i?" %d":"%d",ans[i]);puts("");}return 0;
}
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