poj 2828 Buy Tickets

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_i in 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

这道题用线段树做，看了别人博客上的思路后我自己做出来了，其实这道题和逆序对那道题差不多，就是从后往前插入，因为最后一个的位置一定是确定的，所以从最后一个开始确定位置。确定的时候可以建一颗线段树，用pos储存当前这点下剩余的位置，value表示这一点的价值，这个只当b[i].l==b[i].r的时候才用，相当于是根节点。每次输入一个节点，看看左子树的pos是不是比所插入所要的位置大，如果是那么直接插入，如果不是就往右子树走，并且这时的位置要减去左子树有的位置，即pos-b[i*2].pos.

#include<stdio.h> #include<string.h> #define maxn 200006 int a[maxn],c[maxn],d[maxn]; struct node { int pos,value,l,r; }b[4*maxn]; void build(int l,int r,int i) { int mid; b[i].l=l;b[i].r=r; if(l==r){ b[i].pos=1;return; } mid=(l+r)/2; build(l,mid,i*2); build(mid+1,r,i*2+1); b[i].pos=b[i*2].pos+b[i*2+1].pos; } void update(int pos,int value,int i) { if(b[i].l==b[i].r){ b[i].value=d[b[i].l]=value;b[i].pos=0; return; } b[i].pos--; if(b[i*2].pos>pos) update(pos,value,i*2); else { update(pos-b[i*2].pos,value,i*2+1); } } int main() { int n,m,i,j,T; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++){ scanf("%d%d",&a[i],&c[i]); } build(1,maxn,1); for(i=n;i>=1;i--){ update(a[i],c[i],1); } for(i=1;i<=n;i++){ printf("%d ",d[i]); } printf(" "); } return 0; }