poj 2481 Cows

http://poj.org/problem?id=2481

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 <= 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 cow_i. 

Sample Input

3
1 2
0 3
3 4
0

Sample Output

1 0 0

题意是给出了一堆牛的能力值，这些能力值是用一定区间表示的。一头牛比另一头牛强壮就是这头牛的能力值区间真包含了另一头牛的能力值区间(能力值区间相等不能算作比其强壮)。问对于每一头牛来说，在这个群体中有多少头牛比自己强壮。

http://www.cnblogs.com/111qqz/p/4700882.html

009	class TNode implements Comparable{

010	int s;//左端点

011	int e;//右端点

012	int label;//牛的序号

013	public int compareTo(Object o) {

014	int v=((TNode)o).e;

015	if(this.e!=v) //按右端点降序排列

016	return v-this.e;

017	return this.s-((TNode)o).s;//右端点相等,按左端点升序排序

018

}

019

020	public String toString(){

021	return ("["+s+","+e+"]");

022

}

023

}

024

025	public class Main{

026	static  int N=100015;

027	TNode[] cow;

028	int cal[];//树状数组

029	int res[];

030

int maxn;

031

032	public Main(){

033

034

}

035

036

037

038	private int lowbit(int t){//计算cal[t]展开的项数

039	return t&(-t);

040

}

041

042	private int Sum(int k){ //求前k项的和.

043	int sum=0;

044	while(k>0){

045	sum+=cal[k];

046	k=k-lowbit(k);

047

}

048	return sum;

049

}

050

051	private void update(int i,int x){    //增加某个元素的大小,给某个节点 i 加上 x

052	while(i<=maxn){

053	cal[i]=cal[i]+x; //更新父节点

054	i=i+lowbit(i);

055

}

056

}

061	public static void  main(String args[]) throws IOException{

062	Main ma=new Main();

063

ma.go();

064

}

066	public void go() throws IOException{

067

068	StreamTokenizer st = new StreamTokenizer(new BufferedReader(

069	new InputStreamReader(System.in)));

070	PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

071

072	while(true) {

073	st.nextToken();

074	int n= (int) st.nval;    //牛的个数

075	if(n==0) break;

076

077	cow=new TNode[N];

078	cal=new int[N];

079	res=new int[N];

080	for(int i=0;i< N;i++){

081	cow[i]=new TNode();

082	// cal[i]=0;

083

}

084	for(int i=0;i< n;i++) {//读入牛的吃草区间

085	st.nextToken();

086	cow[i].s=(int) st.nval;

087	st.nextToken();

088	cow[i].e=(int) st.nval;

089	cow[i].s++;

090	cow[i].e++;

091	cow[i].label=i;//牛的原始序号

092	if(cow[i].e>maxn) maxn=cow[i].e;//最大右端点

093

}

094

095	Arrays.sort(cow);//排序

096

097	// for(int i=0;i< n;i++)

098	// System.out.println(cow[i]);

099

100	for(int i=0;i< n;i++) {

101	if(i!=0&&cow[i].s==cow[i-1].s&&cow[i].e==cow[i-1].e)//两头牛有相同的吃草区间

102	res[cow[i].label]=res[cow[i-1].label];//它们有相同的答案

103	else res[cow[i].label]=Sum(cow[i].s);//统计比cow[i].label这头牛强的牛的数目

104	update(cow[i].s,1);//更新

105

}

106	System.out.printf("%d",res[0]);

107	for(int i=1;i< n;i++) System.out.printf(" %d",res[i]);

108	System.out.println();

109

}

110

}