ACM专题学习|Buy Tickets--线段树（单点修改） c语言|c++|算法

ACM专题学习五
题目 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 input
77 33 69 51
31492 20523 3890 19243
题意小猫春节买票，排队无聊要我们做题（？？？），有n个人，编号是val，要排的位置没人就排，有人就进行插队，其序号变为pos，其后面的人位置都后退。
分析如果没做过类似题，其实不好想到用线段树做。建立线段树，每个节点的值为下分支的和。最后一个进入队伍的人的位置就是ta想去的位置，位置是可以确定的，所以我们可以倒序插入，然后记录插入过的位置（把插入过的位置值改为0）。插入的时候比较插入点与节点一半的值，大于等于就走右边，小于就走左边，走一次就用位置减去左分支的值，这样找大树中的位置相当于找小树中的位置。
树的最底层即叶子，总左往右就是队伍的顺序，每一次修改都用数组pri[l]记录val值，每次修改完都更新树的信息（可以自行画图演示下）
代码

#include #include #define MID (l + r) >> 1 #define MAX 200010 using namespace std; int pri[MAX]; int cnt[MAX << 2]; int pos[MAX],val[MAX]; inline void gx(int rt) { cnt[rt] = cnt[rt << 1] + cnt[rt*2+1]; } void build(int l,int r,int rt) { if(l == r) { cnt[rt] = 1; return ; } int m = MID; build(l,m,rt << 1); build(m + 1,r,rt*2+1); gx(rt); } void update(int pos,int val,int l,int r,int rt) { if(l == r) { cnt[rt] = 0; pri[l] = val; return ; } int m = MID; if(cnt[rt << 1] > pos) update(pos,val,l,m,rt << 1); else update(pos - cnt[rt << 1],val,m + 1,r,rt*2+1); gx(rt); } int main() { int n; while(~scanf("%d",&n)) { build(0,n - 1,1); for(int i = 0; i < n; i++) scanf("%d%d",&pos[i],&val[i]); for(int i = n - 1; i >= 0; i--) update(pos[i],val[i],0,n - 1,1); for(int i = 0; i < n; i++) printf("%d%c",pri[i],i < n - 1 ? ' ' : '\n'); } return 0; }