poj 2481 Cows

// 题意:有n头奶牛,各自有属性[s,e],假如Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, 则牛 i 比牛 j 强壮
// 询问每头奶牛比他强的奶牛的数量

#include <iostream>        //一维树状数组
#include <algorithm>
using namespace std;
const int maxn=100005;
int table[maxn],tail;    //table表示树状数组,下标从 1 到 tail
int overlap[maxn],ans[maxn];        

struct Cow 
{
    int st,ed;
    int pos;    //标记在原序列中的位置
    bool operator<(const Cow& c) const        //ed按降序排，若相同则按st升序排  
    {
        if(ed==c.ed)
            return st<c.st;
        else
            return ed>c.ed;
    }
}cow[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void modify(int x)
{
    while(x<=tail)
    {
        table[x]++;        //计数加 1
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int s=0;
    while(x>0)
    {
        s+=table[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        tail=0;
        int i;
        for(i=0;i<n;++i)
        {
            cow[i].pos=i;
            scanf("%d%d",&cow[i].st,&cow[i].ed);    
            cow[i].st++;cow[i].ed++;        //让数组下标从 1开始 
            tail=max(tail,cow[i].st);
        }
        sort(cow,cow+n);
        for(i=1;i<n;++i)    //标记重叠区间
        {
            if(cow[i].ed==cow[i-1].ed&&cow[i].st==cow[i-1].st)
                overlap[i]=overlap[i-1]+1;
            else
                overlap[i]=0;
        }
        memset(table,0,sizeof(table));
        for(i=0;i<n;++i)    
        {
            ans[cow[i].pos]=sum(cow[i].st)-overlap[i];        //找出比当前cow更壮的cow的数目
            modify(cow[i].st);        //更新数据
        }
        for(i=0;i<n;++i)
            printf("%d ",ans[i]);
        printf("\n");
    }
    return 0;
}