题意

输入队伍长度n
接下来n行，a，b 表示b插在队伍的a处
求队伍最后的情况

题解

刚开始并不知道要用线段树，经大佬点悟，发现最后插入的位置就是对应的a。所以可以从后往前依次插入，每次的位置pos即找到一个位置pos使它前面空出pos个位置。然后我们就可以用线段树了。
常数巨大的丑陋代码

# include <stdio.h>

# include <stdlib.h>

# include <iostream>

# include <string.h>

# include <algorithm>

using namespace std;

# define IL inline

# define RG register

# define UN unsigned

# define ll long long

# define rep(i, a, b) for(RG int i = a; i <= b; i++)

# define per(i, a, b) for(RG int i = b; i >= a; i--)

# define uev(e, u) for(RG int e = ft[u]; e != -1; e = edge[e].nt)

# define mem(a, b) memset(a, b, sizeof(a))

# define max(a, b) ((a) > (b)) ? (a) : (b)

# define min(a, b) ((a) < (b)) ? (a) : (b)

const int MAXN = 200001, MAXM = 1000001;

struct Tree{

    int l, r, num;

} tree[MAXM];

int n, ans, pos[MAXN], val[MAXN], a[MAXN];

IL void Updata(RG int now){

    tree[now].num = tree[now<<1].num+tree[now<<1|1].num;

}

IL void Build(RG int now, RG int l, RG int r){

    tree[now] = (Tree) {l, r, 1};

    if(l == r) return;

    RG int mid = l+r>>1;

    Build(now<<1, l, mid); Build(now<<1|1, mid+1, r);

    Updata(now);

}

IL int Query(RG int now, RG int x){

    if(tree[now].l == tree[now].r) return tree[now].r;

    if(x <= tree[now<<1].num) return Query(now<<1, x);

    else return Query(now<<1|1, x-tree[now<<1].num);

}

IL void Add(RG int now, RG int x){

    if(tree[now].l == tree[now].r){

        tree[now].num = 0;

        return;

    }

    RG int mid = tree[now].l+tree[now].r>>1;

    if(mid < x) Add(now<<1|1, x);

    if(mid >= x) Add(now<<1, x);

    Updata(now);

}

int main(){

    while(~scanf("%d", &n)){

        Build(1, 1, n);

        rep(i, 1, n) scanf("%d%d", &pos[i], &val[i]), pos[i]++;

        per(i, 1, n){

            RG int t = Query(1, pos[i]);

            a[t] = val[i];

            Add(1, t);

        }

        rep(i, 1, n) printf("%d ", a[i]);

        printf("\n");

    }

    return 0;

}

scanf前没写~居然TLE了三次