POJ 2481 Cows 树状数组

最新推荐文章于 2019-08-29 15:36:39 发布

HQD因为有趣所以做题

最新推荐文章于 2019-08-29 15:36:39 发布

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分类专栏： ACM_数据结构文章标签： tree mathematica input output each distance

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本文介绍了一道关于处理一维数轴上多个区间查询的问题，通过使用线段树和离线查询的方法，有效地解决了每头牛相对于其他牛的强弱比较问题。

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Cows

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 4786		Accepted: 1478

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 <= 10 ⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 ⁵) 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

Sample Output

1 0 0

Hint

Huge input and output,scanf and printf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct t
{
    int x,y,num,value;
}e[100003];
int tree[100003];
bool cmp(t a,t b)
{
    if(a.y!=b.y) return a.y<b.y;
    return a.x<b.x;
}
bool cmp1(t a,t b)
{
    return a.num<b.num;
}
inline int Lowbit(int x)
{
    return x&(-x);
}
void Update(int x)
{
    for(int i=x;i<100002;i+=Lowbit(i))
         tree[i]++;
}
int Getsum(int x)
{
    int temp=0;
    for(int i=x;i>0;i-=Lowbit(i))
         temp+=tree[i];
    return temp;
}
int main()
{
    int n;
    while(((scanf("%d",&n),n),n)!=0)
    {
        memset(tree,0,sizeof(tree));
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&e[i].x,&e[i].y);
            e[i].y=100001-e[i].y;
            e[i].num=i;
            e[i].value=0;
        }
        sort(e+1,e+1+n,cmp);
        int k=0;
        for(int i=1;i<=n;i++)
        {
            e[i].value=Getsum(e[i].x+1);
            if(i>1&&e[i].x==e[i-1].x&&e[i].y==e[i-1].y)
            {
                k++;
                e[i].value-=k;
            }
            else
                k=0;
            Update(e[i].x+1);
        }
        sort(e+1,e+1+n,cmp1);
        for(int i=1;i<n;i++)
        printf("%d ",e[i].value);
        printf("%d/n",e[n].value);
    }
    return 0;
}