POJ 2828 Buy Tickets(线段树—查找并更新从左到右的第i个1)
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 17243 | Accepted: 8557 |
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 Posi and Valiin 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
40 771 512 691 33 41 192430 20523 1 38900 31492
Sample Output
77 33 69 5131492 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.
题意：队伍里有n个人，这些人是按序到达的，但是他们乱插队。 每个人有两个值pos和val，pos表示这个人会插队到第pos个人后面去，（第0个人是售票窗口）。 插队完毕后，按队伍的顺序输出每个人的val值。
题解：这一题我们需要逆序插入。我们先建立一棵叶子节点全部都为1的线段树（r位置为1表示r位置的位置还有一个空位，将目标插入r位置后，区间[r，r]变为0，表示这个位置已经被占用了。同样当i表示区间[l，r]时，sum[i]=3就表示在区间[l，r]中还有三个空位置）。 当我们逆序插入时，此题就变成了查找并更新从左到右的第pos+1个1。 在更新中，我们记录下每个节点插入的val值，并最后按照从左往右的顺序输出。
代码如下：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 200020
#define lson i*2,l,m
#define rson i*2+1,m+1,r
int pos[maxn],val[maxn];
int val_prin[maxn*4],sum[maxn*4];
int cnt,n;
void PushUp(int i)
{
sum[i]=sum[i*2]+sum[i*2+1];
}
void build(int i,int l,int r)
{
if(l==r)
{
sum[i]=1;
return ;
}
int m=(l+r)/2;
build(lson);
build(rson);
PushUp(i);
}
void update(int p,int v,int i,int l,int r)
{
if(l==r)
{
if(sum[i]==1)
{
sum[i]=0;
val_prin[i]=v;
}
return ;
}
int m=(l+r)/2;
if(p<=sum[i*2])
update(p,v,lson);
else
update(p-sum[i*2],v,rson);//往右边查找时，注意要将pos的值减去左边树上的1的个数
PushUp(i);
}
void print(int i,int l,int r)
{
if(l==r)
{
cnt++;
printf("%d",val_prin[i]);
if(cnt<n)
printf(" ");
else
printf("\n");
return ;
}
int m=(l+r)/2;
print(lson);//输出要求是按队伍顺序输出，即在树的节点上是从左往右输出
print(rson);
}
int main()
{
int i;
while(scanf("%d",&n)!=EOF&&n)
{
cnt=0;
build(1,1,n);
for(i=1;i<=n;++i)
scanf("%d%d",&pos[i],&val[i]);
for(i=n;i>0;--i)
{
int p=pos[i];
p++;
int v=val[i];
update(p,v,1,1,n);
}
print(1,1,n);
}
return 0;
}