POJ 题目2481 Cows（树状数组）

最新推荐文章于 2021-05-24 06:28:51 发布

Jogging_Clown

最新推荐文章于 2021-05-24 06:28:51 发布

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分类专栏：树状数组

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本文介绍了一种算法，用于解决牛群中不同牛之间的强弱比较问题。具体而言，对于每头牛，该算法可以确定有多少头牛比它更强。通过使用区间范围和特殊的数据结构，该算法有效地解决了问题。

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Cows

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 13920		Accepted: 4607

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

题目大意：

给出n头牛，他们有两个属性x和y，当A牛的x比B牛的x小于或等于，A牛的y比B牛的y大于或等于，但x和y不能同时等于，则称为A牛比B牛更强大

问对每头牛，有多少头牛比他强大

ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct ss
{
	int e,s,id;
}b[100100];
int val[100100],a[100100];
int cmp(const void *a,const void *b)
{
	if((*(struct ss *)a).s==(*(struct ss *)b).s)
		return (*(struct ss *)a).e-(*(struct ss *)b).e;
	return (*(struct ss *)b).s-(*(struct ss *)a).s;
}
int n;
int lowbit(int x)
{
	return x&(-x);
}
void update(int p,int q)
{
	while(p<=n)
	{
		a[p]+=q;
		p+=lowbit(p);
	}
}
int sum(int p)
{
	int ans=0;
	while(p>0)
	{
		ans+=a[p];
		p-=lowbit(p);
	}
	return ans;
}
int main()
{
	//int n;
	while(scanf("%d",&n)!=EOF,n)
	{
		int i;
		memset(val,0,sizeof(val));
		memset(a,0,sizeof(a));
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&b[i].e,&b[i].s);
			b[i].id=i;
			b[i].e++;
			b[i].s++;
		}
		qsort(b+1,n,sizeof(b[0]),cmp);
		val[b[1].id]=sum(b[1].e);
		update(b[1].e,1);
		for(i=2;i<=n;i++)
		{
			if(b[i].e==b[i-1].e&&b[i].s==b[i-1].s)
			{
				val[b[i].id]=val[b[i-1].id];
			}
			else
			{
				val[b[i].id]=sum(b[i].e);
			}
				update(b[i].e,1);
		}
		printf("%d",val[1]);
		for(i=2;i<=n;i++)
		{
			printf(" %d",val[i]);
		}
		printf("\n");
	}
}