poj2828(线段树单点更新)

Buy Tickets

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 Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:

Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-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.
Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
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

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题意：有很多人排队，其中有插队的，每个人都有一个值。现在给出这些人排队插队的位置以及他们的值，要求输出最后形成的队列

思路：每一次插队就像线段树里面的单点更新，只不过需要考虑插入位置。而插队的肯定在被插队的前面并且最后插队的人的位置是不变的，所以建一颗树来维护当前区间空位，然后倒着考虑进行单点更新从最后一个插队到第一个在这里排队的人，这样每一次插入都会在插队后面，最后形成整个队列

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <cmath>#include <algorithm>#include <functional>#define inf 0X3f3f3f3fusing namespace std;typedef long long ll;const int MAXN=1e9+10;const int MAX=200000+10;int tree[MAX*4];int ans[MAX*4];int s[MAX],num[MAX],temp=0;int n;void pushup(int node){    tree[node]=tree[node<<1]+tree[node<<1|1];}void init(int node,int l,int r){    tree[node]=(r-l)+1;    if(l==r)    return ;    int mid=(l+r)>>1;    init(node<<1,l,mid);    init(node<<1|1,mid+1,r);    //pushup(node);}void updata_node(int node,int l,int r,int locat,int flag){    if(l==r){        ans[node]=flag;        tree[node]--;        return ;    }    int mid=(l+r)>>1;    if(tree[node<<1]>=locat)        updata_node(node<<1,l,mid,locat,flag);    else        updata_node(node<<1|1,mid+1,r,locat-tree[node<<1],flag);    pushup(node);}void solve(int node,int l,int r){    if(l==r){        if(!temp){            cout<<ans[node];            temp++;        }        else{            cout<<" "<<ans[node];        }        return ;    }    int mid=(l+r)>>1;    solve(node<<1,l,mid);    solve(node<<1|1,mid+1,r);}int main(){    #ifdef SIYU    freopen("in.txt","r",stdin);    #endif // SIYU    while(scanf("%d",&n)!=EOF){        temp=0;        init(1,1,n);        for(int i=1;i<=n;i++){            scanf("%d%d",&s[i],&num[i]);        }        for(int i=n;i>=1;i--){            updata_node(1,1,n,s[i]+1,num[i]);        }        solve(1,1,n);        cout<<endl;    }    return 0;}