POJ2828 Buy Tickeys 树状数组+二分 线段树留坑。。。

题意：一堆人排队买票，告诉你一堆人的序号（序号的意思是插在“当前”第几个人的后面）和姓名（姓名用编号代替）。
思路：线段树 或 树状数组+二分（自己还不会线段树，所以这里继续留坑）。
首先想到，越晚入队的人，它的位置就越正确。比如说最后一个入队的人，插在第8个人后面，那在最终结果里，他一定是第9个人！所以我们考虑倒序处理数据。这题思路和POJ2182一样了，都是倒序树状数组处理+二分查找。每当找到那个人的位置，就在树状数组中把它减去，然后再找下一个就行了。下面是AC代码。
另外一个坑点是，题目要求是2e5，但是就是re，开到2e6就A了。。。莫名其妙我也不知道咋回事。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define cl(a,b) memset(a,b,sizeof(a))
#define lowbit(x) (x&(-x))
using namespace std;
const int maxn=2e6+10;
int n;
int a[maxn],b[maxn],q[maxn],c[maxn];//ab存初始信息，q存最终序列  c存树状数组
void add(int i,int x) {
	while(i<=n) {
		c[i]+=x;
		i+=lowbit(i);
	}
}
int sum(int i) {
	int ans=0;
	while(i>0) {
		ans+=c[i];
		i-=lowbit(i);
	}
	return ans;
}
int find(int v) {
	int l=1,r=n;
	int ans;
	while(l<=r) {
		int mid=(l+r)>>1;
		int temp=sum(mid);
		if(temp<v) {
			l=mid+1;
		}
		if(temp>=v) {
			ans=mid;
			r=mid-1;
		}
	}
	return ans;
}
int main() {
	while(scanf("%d",&n)!=EOF) {
		for(int i=1; i<=n; i++) {
			scanf("%d%d",&a[i],&b[i]);
		}
		for(int i=1; i<=n; i++) {
			add(i,1);
		}
		for(int i=n; i>=1; i--) {
			int p=find(a[i]+1);
			add(p,-1);
			q[p]=b[i];
		}
		for(int i=1; i<n; i++) {
			printf("%d ",q[i]);
		}
		printf("%d\n",q[n]);
	}
	return 0;
}

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
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.