POJ 2828 Buy Tickets

Buy TicketsTime Limit: 4000 MS Memory Limit: 65536 KTotal Submissions: 10858 Accepted: 5272 DescriptionRailway 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 Olympus in ICS ICs. 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 shi Ver. 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 wocould 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. inputThere 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 ε [0, I −1]-The I-th person came to th E 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 ε [0, 32767]-The I-th person was assigned the value Vali. there no blank lines between test cases. proceed to the end of input. outputFor each test cases, output a single line of space-separated I Ntegers which are the values of people in the order they stand in the queue. sample Input40 771 511 332 6940 205231 192431 38900 Sample Output77 33 69 31492 5131492 20523 HintThe 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. sourcePOJ Monthly -- 2006.05.28, Zhu, Zeyuan was depressed for many days and finally couldn't help looking at the solution report, and then...

#include <iostream>  #include <cstring>  #include <cstdio>  #include <cmath>  #define N 200010  using namespace std;  struct num  {      int l,r,sum;  }a[4*N];  struct inf  {      int sum,val;  }in[N];  int sa[N],res;  int main()  {      //freopen("data.in","r",stdin);      void pre_build(int k,int l,int r);      void find(int k,int sum);      int n;      while(scanf("%d",&n)!=EOF)      {          for(int i=1;i<=n;i++)          {              scanf("%d %d",&in[i].sum,&in[i].val);          }          pre_build(1,1,n);          for(int i=n;i>=1;i--)          {              find(1,in[i].sum+1);              sa[res] = i;          }          for(int i=1;i<=n;i++)          {              if(i==1)              {                  printf("%d",in[sa[i]].val);              }else              {                  printf(" %d",in[sa[i]].val);              }          }          printf("\n");      }      return 0;  }  void pushup(int k)  {      a[k].sum=a[k<<1].sum+a[k<<1|1].sum;  }  void pre_build(int k,int l,int r)  {      a[k].l=l; a[k].r = r;      if(l==r)      {          a[k].sum=1;          return ;      }      int mid=(l+r)>>1;      pre_build(k<<1,l,mid);      pre_build(k<<1|1,mid+1,r);      pushup(k);  }  void find(int k,int sum)  {      if(a[k].l==a[k].r)      {          res = a[k].l;          a[k].sum=0;          return ;      }      if(a[k<<1].sum>=sum)      {          find(k<<1,sum);      }else      {          find(k<<1|1,sum-a[k<<1].sum);      }      pushup(k);  }