Cows

Time Limit: 3000MS

 

Memory Limit: 65536K

Total Submissions: 23460

 

Accepted: 7898

Description

Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. 

Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E]. 

But some cows are strong and some are weak. Given two cows: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj. 

For each cow, how many cows are stronger than her? Farmer John needs your help!

Input

The input contains multiple test cases. 
For each test case, the first line is an integer N (1 <= N <= 105), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 105) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge. 

The end of the input contains a single 0.

Output

For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cowi. 

Sample Input


3 1 2 0 3 3 4 0


Sample Output


1 0 0


Hint

Huge input and output,scanf and printf is recommended.

Source

​​POJ Contest​​,Author:Mathematica@ZSU

 

题意：求一个区间真子集的个数。

分析：数组数组只能实现单点查询，所以这题需要转化一下

首先，我们先对区间进行排序，按照e的位置从大到小排序，e相同，按照s，从小到大排序，数组数组标记s。

如果相同的话不算真子集，所以需要判断一下，直接令当前等于前一个即可。

这题核心就是排序后的数后面的不可能比前面的强壮。

#include<stdio.h>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int MAXN=200000+5;//最大元素个数
int n;//元素个数
ll c[MAXN],ans[MAXN];//c[i]==A[i]+A[i-1]+...+A[i-lowbit(i)+1]
struct node{
    int s,e,index;
}a[MAXN];
//返回i的二进制最右边1的值
int lowbit(int i)
{
    return i&(-i);
}
//返回A[1]+...A[i]的和
ll sum(int x){
    ll sum = 0;
    while(x){
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}
//令A[i] += val
void add(int x, ll val){
    
    while(x <= n){
        c[x] += val;
        x += lowbit(x);
    }
}
bool cmp(const node&a,const node &b)
{
    if(a.e!=b.e)
        return a.e>b.e;
    else
        return a.s<b.s;
} 
int main()
{
    while(scanf("%d",&n)!=-1&&n)
    {
        
        memset(c,0,sizeof(c));
        memset(ans,0,sizeof(ans));
         for(int i=1;i<=n;i++)
         {
           scanf("%d%d",&a[i].s,&a[i].e);
           a[i].s++;  ///数组数组保存下标从1开始
           a[i].e++;
            a[i].index=i;
         }
         sort(a+1,a+n+1,cmp);
         ans[a[1].index]=0;
         add(a[1].s,1);
         for(int i=2;i<=n;i++)
         {
            if(a[i].s==a[i-1].s&&a[i].e==a[i-1].e)
                ans[a[i].index]=ans[a[i-1].index];
            else
                ans[a[i].index]=sum(a[i].s);
                
            add(a[i].s,1);
         
         }
         for(int i=1;i<n;i++)
         {
            printf("%d ",ans[i]);
         }
         cout<<ans[n]<<endl;
    }
    return 0;
}