NOJ 1317——Exercise of Sort

• 问题描述
• Wodex opened the door successfully. 
  At the same time, horn sounded: 
  If you are not clever, the game will be so boring, next level will be more difficult. Ha-ha-ha... 
  
  So, Wodex came in with his partners. 
  At this room, there was a ciphertext on the wall. 
  The words means: 
  There are N lines, each line has two integers i and x, it means value x will be put on ith position. Every x is different. 
  
  How to form the sequence of these values? 
  
  
• 输入
• Input until EOF. 
  First line will contain an integer N (1 <= N <= 200,000), then N lines follow, each line has two integers i (0 <= i < N) and x (0 <= x <= 1,000,000). 
  
• 输出
• Output the final sequence of these numbers. 
  
• 样例输入

• 4
  0 77
  1 51
  1 33
  2 69
  4
  0 20523
  1 19243
  1 3890
  0 31492

• 样例输出

• 77 33 69 51
  31492 20523 3890 19243

• 提示

• 无

• 来源

• Hungar

线段树完成，最后一个数的位置一定是确定的。具体见http://hainingqtj.blog.163.com/blog/static/23349311920146411520361/

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<iostream>

using namespace std;

const int maxn = 200010;

int tree[maxn<<2];

int _sort[maxn];

struct node
{
    int val,pos;
}num[maxn];

void build(int p,int l,int r)
{
    tree[p]=r-l+1;
    if(l == r)
        return;
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
}

void update(int p,int l,int r,int val,int pos)
{
    tree[p]--;
    if(l == r)
    {
        _sort[l]=val;
        return ;
    }
    int mid=(l+r)>>1;
    if(tree[p<<1] >= pos)
        update(p<<1,l,mid,val,pos);
    else
    {
        pos-=tree[p<<1];
        update(p<<1|1,mid+1,r,val,pos);
    }
        
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        build(1,1,n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&num[i].pos ,&num[i].val);
        }
        for (int i = n-1; i >= 0; --i)
        {
            update(1,1,n,num[i].val,num[i].pos+1);
        }
        for (int i = 1; i <= n; ++i)
        {
            printf("%d",_sort[i]);
            if(i<n)
                printf(" ");
        }
        printf("\n");
    }
    return 0;
}