插入队列进行逆向操作，pos代表的也就是放入此人时前面应有pos个空位，用线段树维护空位数量。
/****************************************************/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <stack>
#include <map>
#include <queue>
#include <algorithm>
#include <ctime>
#define EPS 1E-8
#define MAXN 200010
#define INF (~0U >> 2)
#define MOD 1000000007
#define Lson l, mid, rt << 1
#define Rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long LL;
/****************************************************/
int num[MAXN << 2], res[MAXN], p;
struct P
{
int val, pos;
}a[MAXN];
void push_up(int rt)
{
num[rt] = num[rt << 1] + num[rt << 1 | 1];
}
void build(int l, int r, int rt)
{
if (l == r)
{
num[rt] = 1;
return;
}
int mid = (l + r) >> 1;
build(Lson);
build(Rson);
push_up(rt);
}
void query(int x, int l, int r, int rt)
{
if (l == r)
{
//puts("ok");
p = l;
//cout << p << endl;
num[rt] = 0;
return;
}
int mid = (l + r) >> 1;
if (num[rt << 1] > x)
query(x, Lson);
else
query(x - num[rt << 1], Rson);
push_up(rt);
}
int main()
{
int N;
while (scanf("%d", &N) != EOF)
{
build(1, N, 1);
for (int i = 0; i < N; i++)
{
scanf("%d%d", &a[i].pos, &a[i].val);
}
for (int i = N - 1; i >= 0; i--)
{
int x = a[i].pos;
query(x, 1, N, 1);
res[p] = a[i].val;
}
printf("%d", res[1]);
for (int i = 2; i <= N; i++)
printf(" %d", res[i]);
puts("");
}
}