Cows

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: cow
_i and cow
_j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei – Si > Ej – Sj, we say that cow
_i is stronger than cow
_j.

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 <= 10
⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10
⁵) 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 cow
_i.

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         就是给出N个区间，问这个区间是多少个区间的真子集。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=100010;
int c[MAXN];
struct Node
{
    int S,E;
    int index;
}node[MAXN];


int n;
int cnt[MAXN];//记录结果

//先按照E从大到小排序，E相同则按照S从小到大排序
bool cmp(Node a,Node b)
{
    if(a.E==b.E)return a.S<b.S;
    return a.E>b.E;
}
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&node[i].S,&node[i].E);
            node[i].index=i;
        }
        sort(node+1,node+n+1,cmp);
        memset(c,0,sizeof(c));
        memset(cnt,0,sizeof(cnt));
        cnt[node[1].index]=0;
        add(node[1].S+1,1);
        for(int i=2;i<=n;i++)
        {
            if(node[i].E==node[i-1].E&&node[i].S==node[i-1].S)
               cnt[node[i].index]=cnt[node[i-1].index];
            else
                cnt[node[i].index]=sum(node[i].S+1);
            add(node[i].S+1,1);
        }
        printf("%d",cnt[1]);
        for(int i=2;i<=n;i++)
          printf(" %d",cnt[i]);
        printf("\n");
    }
    return 0;
}