POJ 2828 单点更新（好题）

Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 15086		Accepted: 7530

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 Pos_i and Val_iin the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

• Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Pos_i-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.
• Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

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

 

 

题目意思：

给n个人信息，posi表示插到第i个人后面，vali表示这个人的价值，输出插入完毕后价值序列。

 

思路：

从前往后的话很难写出来，因为后面的人会影响前面人的位置。那么就从后往前，直接用例子解释吧

5

0 1

0 2

1 3

1 4

0 5

 

从后往前，先插入第5个人

插入的时候这个人前面肯定为0个人，那么序列为5 -1 -1 -1 -1（-1表示这个位置为空）

插入第4个人

插入的时候这个人前面肯定有1个人，那么给那个人留一个空位，序列为5 -1 4 -1 -1

插入第3个人

前面肯定有1个人，给那个人留下一个空位，序列为5 -1 4 3 -1

插入第2个人

前面肯定有0个人，不留空位，序列为5 2 4 3 -1

插入第1个人

只能放在最后面了   序列为5 2 4 3 1

 

这样的话，套个线段树就可以了。

 

代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <set>
 using namespace std;

 #define N 2000005
 #define ll root<<1
 #define rr root<<1|1
 #define mid (a[root].l+a[root].r)/2

 int max(int x,int y){return x>y?x:y;}
 int min(int x,int y){return x<y?x:y;}
 int abs(int x,int y){return x<?-x:x;}

 int n;
 int pos[N], val[N];
 int ans[N];

 struct node{
     int l, r, num;
 }a[N];

 void build(int l,int r,int root){
     a[root].l=l;
     a[root].r=r;
     if(l==r) {
         a[root].num=;
         return;
     }
     build(l,mid,ll);
     build(mid+,r,rr);
     a[root].num=a[ll].num+a[rr].num;
 }

 void solve(int id,int root){
     if(a[root].l==a[root].r){
         ans[a[root].l]=val[id];
         a[root].num--;
         return;
     }
     if(a[ll].num>=pos[id]+) solve(id,ll);//前面有pos[id]+1个空位，其中一个是自己的
     else {//否则往后面（右子树）插入
         pos[id]-=a[ll].num;
         solve(id,rr);
     }
     a[root].num=a[ll].num+a[rr].num;
 }

 main()
 {
     int i, j, k;
     while(scanf("%d",&n)==){
         for(i=;i<=n;i++){
             scanf("%d %d",&pos[i],&val[i]);
         //    pos[i]++;
         }
         build(,n,);
         for(i=n;i>=;i--){
             solve(i,);
         }
         printf("%d",ans[]);
         for(i=;i<=n;i++) printf(" %d",ans[i]);
         cout<<endl;
     }
 }