hdu Cows

Problem 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

   
   
   
   

제목은 각 구간이 몇 개의 구간에 포함되어 있는지를 구하는 것이다...

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define maxx 100010
int tree[maxx];
struct Node
{
	int x,y,i;
}node[maxx];
int lowbit(int v)
{
	return v&-v;
}
void add(int v,int var)
{
	while(v<=maxx){
		tree[v]+=var;
		v+=lowbit(v);
	}
}
int cmp(Node a,Node b)
{
	if(a.y==b.y)  return a.x<b.x;// ， ，  
	else return a.y>b.y;
}
int querry(int v)
{
	int sum=0;
	while(v>0){
		sum+=tree[v];
		v-=lowbit(v);
	}
	return sum;	
}
int main()
{
	int t,n,i,j,a,b,c[maxx];
	while(scanf("%d",&n),n){
	memset(tree,0,sizeof(tree));//  
	memset(c,0,sizeof(c));	
	for(i=1;i<=n;i++){
			node[i].i=i;// ， sort  
			scanf("%d%d",&node[i].x,&node[i].y);
		}
	sort(node+1,node+n+1,cmp);
	c[node[1].i]=querry(node[1].x+1);
	add(node[1].x+1,1);
	for(i=2;i<=n;i++){
		if(node[i].x==node[i-1].x && node[i].y==node[i-1].y){//  
		c[node[i].i]=c[node[i-1].i]; 
		add(node[i].x+1,1);
		continue;
		}
		c[node[i].i]=querry(node[i].x+1);// ... 
		add(node[i].x+1,1);
	}
	for(i=1;i<n;i++){
		printf("%d ",c[i]);
	}
	printf("%d
",c[n]);
	}
	return 0;
}