Buy Tickets
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 16010 Accepted: 7983
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.
Source
POJ Monthly–2006.05.28, Zhu, Zeyuan
插队问题,就是每个人都有自己的位置,当自己的位置有人时,那个人就要向后移动,输出对应位置上人的vail,所以对于最后的人,他的位置是固定的,而之前的人的位置是变动的,所以可以从后向前,进行插入.
但我发现一个奇怪的问题,g++超时,c++ AC,将代码的输出改成每个数据后面都有空格g++就可以AC 神题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;
const int MAX = 210000;
int num[MAX*4];//记录num[i]对应的区间还可以插入的位置个数
int w[MAX];
int a[MAX],ww[MAX];
void Build(int L,int R,int site)//初始化
{
if(L==R)
{
num[site]=1;
return ;
}
int mid = (L+R)>>1;
Build(L,mid,site<<1);
Build(mid+1,R,site<<1|1);
num[site]=num[site<<1]+num[site<<1|1];
}
void update(int L,int R,int s,int W,int site)
{
if(L==R)
{
num[site]=0;
w[L]=W;
return ;
}
int mid=(L+R)>>1;
if(num[site<<1]>=s)//如果左子树的位置的个数比它要插入的位置大说明可以插入
{
update(L,mid,s,W,site<<1);
}
else//如果能插入左子树,则插入右子树中,相对右子树它插入的位置为(s-num[site<<1]);
{
update(mid+1,R,s-num[site<<1],W,site<<1|1);
}
num[site]=num[site<<1]+num[site<<1|1];
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%d %d",&a[i],&ww[i]);
}
Build(1,n,1);
for(int i=n;i>=1;i--)//逆序插入
{
update(1,n,a[i]+1,ww[i],1);
}
for(int i=1;i<=n;i++)//输出
{
if(i!=1)
{
printf(" ");
}
printf("%d",w[i]);
}
printf("
");
}
return 0;
}