zoj 3635 线段树

最新推荐文章于 2018-02-25 23:01:02 发布

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最新推荐文章于 2018-02-25 23:01:02 发布

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分类专栏：线段树

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链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3635

Cinema in Akiba (CIA) is a small but very popular cinema in Akihabara. Every night the cinema is full of people. The layout ofCIA is very interesting, as there is only one row so that every audience can enjoy the wonderful movies without any annoyance by other audiences sitting in front of him/her.

The ticket for CIA is strange, too. There are n seats in CIA and they are numbered from 1 to n in order. Apparently, n tickets will be sold everyday. When buying a ticket, if there are k tickets left, your ticket number will be an integeri (1 ≤ i ≤ k), and you should choose the ith empty seat (not occupied by others) and sit down for the film.

On November, 11th, n geeks go to CIA to celebrate their anual festival. The ticket number of theith geek is a_i. Can you help them find out their seat numbers?

Input

The input contains multiple test cases. Process to end of file.
The first line of each case is an integer n (1 ≤ n ≤ 50000), the number of geeks as well as the number of seats inCIA. Then follows a line containing n integers a₁,a₂, ..., a_n (1 ≤ a_i ≤n - i + 1), as described above. The third line is an integerm (1 ≤ m ≤ 3000), the number of queries, and the next line ism integers, q₁, q₂, ..., q_m (1 ≤ q_i ≤ n), each represents the geek's number and you should help him find his seat.

Output

For each test case, print m integers in a line, seperated by one space. Theith integer is the seat number of the q_ith geek.

Sample Input

Sample Output

1 2 3
4 5 3 1 2

题意分析:

有n个座位,编号为1~n,每次当有人取了票后，还剩k张票，那么根据空位的相对位置求每个人最终坐的座位编号.

用线段树区间保存区间内空座位的个数，然后每次查询座位时，根据相对空位的位置与之比较，找出座位编号.

Ac code:

#include<stdio.h>
#include<iostream>
#define N 50001
using namespace std;
int f[N];    //记录相对位置的座位编号
int b[N]; 
struct segeTree
{
int left,right;
int account;
}Tree[N*4];
void build(int l,int r,int k)
{
if(l==r)
{
Tree[k].left=l;
Tree[k].right=r;
Tree[k].account=1;
}
else
{
 int mid=(l+r)/2;
 Tree[k].left=l;
 Tree[k].right=r;
 Tree[k].account=r-l+1;
 build(l,mid,k<<1);
 build(mid+1,r,(k<<1)|1);
}
}
int update(int i,int k)
{
int temp;
if(Tree[k].left==Tree[k].right)
{
Tree[k].account=0;    //此位置已被占，故空位位置的个数为0 
        return Tree[k].left;
}
else
{
int mid=Tree[k<<1].account;
if(i<=mid)
temp=update(i,k<<1);
else
temp=update(i-mid,(k<<1)|1);
}
Tree[k].account=Tree[2*k].account+Tree[2*k+1].account; //从下往上更新空位位置个数
return temp;
}
int main()
{
int num,i,j,query;
while(scanf("%d",&num)!=EOF)
{
build(1,num,1);
for(i=1;i<=num;++i)
{
scanf("%d",&j);
f[i]=update(j,1);
}
scanf("%d",&query);
   for(j=1;j<=query;++j)
   {
scanf("%d",&b[j]);
}
for(j=1;j<query;++j)
printf("%d ",f[b[j]]);
printf("%d\n",f[b[j]]);
}
}