Buy
Tickets Time Limit:4000MS
Memory Limit:65536KB 64bit IO
Format:%I64d & %I64u
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 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 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.
【题目大意】；有n个人买火车票，期初没有人排队，然后陆续的来人，来一个人就给他分配一个数字，而且会有人插队，要你输出最后的队伍顺序。
【题目分析】：这是一个经典的逆推问题。需要注意到，最后一个插入的值位置的固定的，也就是最后一个插入的值可以在它想要插入的位置，所以，先保存所以整数对，之后在倒序插入序列中。那到底应该用什么数据结构实现呢？这里我们发现，可以用线段树来解决这一问题，每一个区间添加一个域
num，表示该区间还能放置的位置数，或者说，表示该区间还能放置的最大的序号值（当然，这是相对于最左边的区间而已的，所以，右区间的话，中间还有一步维护操作。
source code:
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#define MAX 200010
using namespace std;
struct Node
{
int p,v;
} node[MAX];
struct TNode
{
int l,r;
int c;
} tree[4*MAX];
int N;
int ans[MAX];
void build(int l,int r,int x)
{
tree[x].c=r-l+1; //该区间的大小,开始写反了，wa了两次
tree[x].l=l;
tree[x].r=r;
if(l==r) //到底了，返回
return;
int mid=(l+r)>>1;
build(l,mid,2*x);
build(mid+1,r,2*x+1);
}void Insert(int x,int pos,int val) //下标 位置(position) 价值(value)
{
tree[x].c--; //该区间的空格数减少1
if(tree[x].l==tree[x].r) //直到找到了一个合适的空格来存放该位置
{
ans[tree[x].l]=val; //将价值映射到ans中
return;
}
if(tree[2*x].c>=pos) //该区间的剩余空格数大于位置的坐标，往前探寻
Insert(2*x,pos,val);
else //该区间剩余空格数小于位置的坐标，往后找空格
Insert(2*x+1,pos-tree[2*x].c,val);
}int main()
{
while(~scanf("%d",&N))
{
build(1,N,1);
int i;
for(i=1;i<=N;i++)
{
scanf("%d%d",&node[i].p,&node[i].v);
node[i].p++; //输入是从0开始的，加1变为从1开始
}
for(i=N;i>=1;i--) //重点：逆序更新
{
Insert(1,node[i].p,node[i].v);
}
for(i=1;i<N;i++)
printf("%d ",ans[i]);
printf("%d\n",ans[N]);
}
return 0;
}
线段树 --- (区间维护+逆推),布布扣,bubuko.com