POJ 2828 Buy Tickets（线段树）

Buy Tickets

Time Limit: 4000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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_iare 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.

题目大意：

n个人站队（可以插队），Pi代表i插队时他前面的人数，Vi代表i的标记，最终输出的是最终队伍的序列。

解题思路：
由于n很大，所以我们不能朴素的挨个添加。我们可以选择用线段树维护，维护的内容是区间内的空位个数。每个位置起初都是1，若该位置有人，则变为0。
如果我们是从前到后依次插队，那么由于后来的人可能会站在已经存在的人的前面，那么我们一次得到的位置不一定是最终位置。所以我们要选择从后往前依次插队，后面的人不会影响到前面的人，若前面的人没有来，将位置空出即可，这样才能才能保证每个人的第一次站队的位置就是最终的位置。

参考代码：

#include<set>#include<map>#include<stack>#include<queue>#include<cmath>#include<vector>#include<cctype>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define lson l,mid,cur<<1#define rson mid+1,r,cur<<1|1#define root 1,n,1const double eps=1e-10;const int INF=0x3f3f3f3f;const int MAXN=200000+50;typedef long long LL;int n,sum[MAXN<<2],loc[MAXN],num[MAXN],people[MAXN];void pushup(int cur){    sum[cur]=sum[cur<<1]+sum[cur<<1|1];}void build(int l,int r,int cur){    if(l==r)    {        sum[cur]=1;        return ;    }    int mid=(l+r)/2;    build(lson);    build(rson);    pushup(cur);}void update(int a,int b,int l,int r,int cur){    if(l==r)    {        people[l]=b;        sum[cur]=0;        return ;    }    int mid=(l+r)/2;    if(sum[cur<<1]>=a)        update(a,b,lson);    else        update(a-sum[cur<<1],b,rson);    pushup(cur);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif // ONLINE_JUDGE    while(scanf("%d",&n)!=EOF)    {        build(root);        for(int i=1;i<=n;i++)            scanf("%d%d",&loc[i],&num[i]);        for(int i=n;i>=1;i--)//从后往前依次插队            update(loc[i]+1,num[i],root);//注意loc[i]要加1        for(int i=1;i<=n;i++)        {            if(i!=n)                printf("%d ",people[i]);            else                printf("%d\n",people[i]);        }    }    return 0;}