POJ 2481 Cows 简单树状数组区间覆盖

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Cows

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 13334 Accepted: 4413

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 <= 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

31 20 33 40

Sample Output

1 0 0

Hint

Huge input and output,scanf and printf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

n头牛，每头牛都有一个范围[s,e]，让比它强壮的牛的个数。第i头牛比第j头牛强壮的定义:si<=sj&&ei>=ej&&ei-si>ej-si。

将n头牛按照e从大到小排序，跟据s的值，用树状数组求得答案。

//2156K1079MS#include<stdio.h>#include<algorithm>#include<string.h>#define M 100007using namespace std;int c[M],ans[M],max_e;struct Cow{    int s,e,id;}p[M];int cmp(Cow a,Cow 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 pos,int val){    while(pos<=max_e+1)    {        c[pos]+=val;        pos+=lowbit(pos);    }}int getsum(int pos){    int res=0;    while(pos>0)    {        res+=c[pos];        pos-=lowbit(pos);    }    return res;}int main(){    int n;    while(scanf("%d",&n),n)    {        max_e=0;        for(int i=0;i<n;i++)        {            scanf("%d%d",&p[i].s,&p[i].e);            p[i].id=i;            if(max_e<p[i].e)max_e=p[i].e;        }        memset(c,0,sizeof(c));        sort(p,p+n,cmp);        for(int i=0;i<n;i++)        {            if(p[i].s==p[i-1].s&&p[i].e==p[i-1].e)ans[p[i].id]=ans[p[i-1].id];//与前面的牛属性一样            else ans[p[i].id]=getsum(p[i].s+1);            add(p[i].s+1,1);        }        for(int i=0;i<n-1;i++)printf("%d ",ans[i]);        printf("%d\n",ans[n-1]);    }    return 0;}