poj 2828--Buy Tickets(线段树）

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.

题意：按顺序给你n个人，给你他想要排的位置pos和他拥有的值val后来着可以插队比如pos[0]=0,pos[1]=0,那么pos[0]就变成了1被插队了。

问最后n个人排好后各自位置是多少，输出各个位置上对应的val值。

举一个例子6个人 1 2 3 4 5 6 位置，倒过来考虑因为最后一个选位置没人和他抢。假设他要2号位那么序列变为 1 0 2 3 4 5前面选2好位的只能退一位。

当前一个人选4号位那么 1 0 2 3 0 4，以此类推排好位置。这种方法用到前缀和就是一开始全部位置初值为1然后求个位置的前缀和，如果这个位置被选了

那么这个位置的值就变成0，更新。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int M = 2e6 + 10;
int re;
int pos[M] , Val[M] , last[M];
struct TnT {
    int l , r , val;
}T[M << 2];
void build(int l , int r , int p) {
    int mid = (l + r) >> 1;
    T[p].l = l , T[p].r = r;
    if(T[p].l == T[p].r) {
        T[p].val = 1;
        return ;
    }
    build(l , mid , p << 1);
    build(mid + 1 , r , (p << 1) | 1);
    T[p].val = T[p << 1].val + T[(p << 1) | 1].val;
}
void updata(int num , int p) {
    if(T[p].l == T[p].r) {
        T[p].val = 0;
        re = T[p].l;
        return ;
    }
    if(num <= T[p << 1].val) {
        updata(num , p << 1);
    }
    else {
        updata(num - T[p << 1].val , (p << 1) | 1);
    }
    T[p].val = T[p << 1].val + T[(p << 1) | 1].val;
}
int main()
{
    int n;
    while(scanf("%d" , &n) != EOF) {
        build(1 , n , 1);
        for(int i = 0 ; i < n ; i++) {
            scanf("%d%d" , &pos[i] , &Val[i]);
            pos[i]++;
        }
        for(int i = n - 1 ; i >= 0 ; i--) {
            updata(pos[i] , 1);
            last[re] = Val[i];
        }
        for(int i = 1 ; i <= n ; i++) {
            printf("%d " , last[i]);
        }
        printf("\n");
    }
    return 0;
}