poj 2481

Language: Default

Cows

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 21941		Accepted: 7419

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

3
1 2
0 3
3 4
0

Sample Output

1 0 0

Hint

Huge input and output,scanf and printf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

简要说明题意：

    给你一些区间，问你每个区间会被多少个区间完全覆盖（并且不完全相同）。

首先根据左端点的位置排序，从小到大访问，在最后把它对应的右端点位置的数+1，只求这个区间r到所有区间的最靠右位置的区间和就是答案，因为我们思考在它前面那些让靠后位置加1的区间，因为排序的原因l一定是小于等于此处的l，那么r也是满足条件就可以计入了

注意相同区间还有 5，6包含 5,5

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+5;
struct A
{
    int l,r,id;
    bool operator<(const A &a)const
    {
        if(a.l!=l) return l<a.l;
        return r>a.r;
    }
}a[N];
int ans[N],c[N],mx;
void add(int x)
{
    for(;x<=mx;x+=x&-x)
        ++c[x];
}
int askk(int x)
{
    int re=0;
    for(;x;x-=x&-x)
        re+=c[x];
    return re;
}
int main()
{
    while(1)
    {
        int n;
        mx=0;
        scanf("%d",&n);
        if(!n) break;
        for(int i=1;i<=n;++i)
            scanf("%d%d",&a[a[i].id=i].l,&a[i].r),mx=max(a[i].r,mx);//取最大右端
        sort(a+1,a+n+1);
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;++i)
        {
            if(a[i].l==a[i-1].l&&a[i].r==a[i-1].r) ans[a[i].id]=ans[a[i-1].id];//与前一个相同则不需要计算
            else ans[a[i].id]=askk(mx)-askk(a[i].r-1);//计算a[i].r到mx的区间和
            add(a[i].r);//对应位置添加
        }
        for(int i=1;i<=n;++i) printf("%d ",ans[i]);
        printf("
");
    }
    return 0;
}