poj 2828POJ2828 Buy Tickets【线段树，逆序遍历】

Buy Tickets

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 19036 Accepted: 9454

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

40 771 511 332 6940 205231 192431 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.

Source

题意： 
排队买票时候插队。 
给出一些数对，分别代表某个人的想要插入的位置Pos_i和他的Val_i，求出最后的队列的val顺序

思路：

只要从后面开始站，假设后面的人都已经站在正确的位置上了，那么到那个人站的时候，现在的位置上已经都是后面的那些人了，只要数posi个空格，那那个人站的位置能确定了。确定之后就可以求下一个了，所以这个前提和结论都成立了。

参考博客http://www.cnblogs.com/CheeseZH/archive/2012/04/29/2476134.html

初始状态

首先是插入3 69

1，4结点有4个位置，

1，2结点有2个位置，小于3，因此放到1，4结点右孩子，且1，4结点空位置减1

到了1，4右孩子后，只要找到第3-2=1个位置即可，而3，4结点的左孩子3，3含有1个空位置，1>=1，所以放到3，3位置了。

 

插入2 33

 

★关键是这里如何处理★

插入2 51

此时1，4的左孩子只有1个位置，1<2，所以只能放到1，4的右孩子3，4上

3，4的左孩子有0个位置，所以只能放在3，4的右孩子4，4上。

 

插入1 77

#include<stdio.h>#include<string.h>#define maxn 200005int tree[maxn<<2];int pos[maxn];//表示前面有pos[i]个空位int val[maxn];//价值int final[maxn];void build(int l,int r,int i){    tree[i] = ( r - l + 1);    if( l == r) return;    int mid = (l+r) >> 1;    build( l ,mid, 2*i);    build( mid + 1, r, 2*i+1);}int update(int p,int l,int r,int v){    tree[v]--;//每插一个空就减少一个    if( l == r)    {        return l;    }    int mid = ( l + r) >> 1;    if( p <= tree[2*v] ) update( p, l, mid, 2*v);//如果左孩子的空位大于所需的空位就向左孩子遍历    else//如果左孩子的空位数小于所需的的空位数    {        p-=tree[2*v];// 那么在右孩子上的空位数等于所需的空位数减去左孩子的空位数        update( p, mid + 1, r, 2*v+1);    }}int main(){    int t;    int i,j;    int n;    while(~ scanf("%d",&n))    {        for(i=1;i<=n;i++)        {            scanf("%d%d",&pos[i],&val[i]);        }        build( 1, n, 1);        for( i = n; i >= 1; i--)        {           int id = update(pos[i] + 1, 1, n, 1);           final[id] = val[i];        }        printf("%d",final[1]);        for( i = 2; i <=n; i++)        {            printf(" %d",final[i]);        }        printf("\n");    }    return 0;}