POJ 2481 Cows 树状数组

题目链接：POJ2481

Cows

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 18579 Accepted: 6249

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.

题意：大意是农场有一些奶牛，每头奶牛有一个喜欢活动的闭区间[S,E];对于奶牛a，b，如果Sa<=Sb&&Eb<=Ea&&Ea-Sa>Eb-Sb就称a比b强壮，问从1到n的n每头奶牛比他强壮的有几头奶牛。

题目分析：可以先将这些奶牛按照E的降序排序，E相同按S的升序，这样一次处理后对于比奶牛i强壮的奶牛都是排序在i的前面并且S值比Si小的牛。使用树状数组统计一下即可，注意区间完全相同的不计数。

////  main.cpp//  Cows////  Created by teddywang on 2017/4/26.//  Copyright © 2017年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>int maxn;using namespace std;struct node {    int x,y;    int pos;    friend bool operator < (node a,node b){        if(a.y==b.y)        {            return a.x<b.x;        }        else return a.y>b.y;    }}s[100010];int N;int x[100010],c[100010],r[100010];int lowbit(int x){    return x&(-x);}int sum(int x){    int ans=0;    while(x!=0)    {        ans+=c[x];        x-=lowbit(x);    }    return ans;}void update(int x,int y){    while(x<=maxn)    {        c[x]+=y;        x+=lowbit(x);    }}int main(){    while(scanf("%d",&N)&&N)    {        memset(c,0,sizeof(c));        memset(x,0,sizeof(x));        maxn=0;        for(int i=0;i<N;i++)        {            scanf("%d%d",&s[i].x,&s[i].y);            s[i].x++;            s[i].pos=i;            maxn=max(maxn,s[i].x);        }        sort(s,s+N);        x[s[0].pos]=0;        update(s[0].x, 1);        for(int i=1;i<N;i++)        {            if(s[i].x==s[i-1].x&&s[i].y==s[i-1].y)                x[s[i].pos]=x[s[i-1].pos];            else            {                x[s[i].pos]=sum(s[i].x);                            }            update(s[i].x,1);        }        for(int i=0;i<N-1;i++)            printf("%d ",x[i]);        printf("%d\n",x[N-1]);    }}