HDOJ 题目3465 Life is a Line（树状数组求逆序对） Life is a Line

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概述

Life is a Line

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1705 Accepted Submission(s): 388

Problem Description

There is a saying: Life is like a line, some people are your parallel lines, while others are destined to meet you.
Maybe have met, maybe just a matter of time, two unparallel lines will always meet in some places, and now a lot of life (i.e. line) are in the same coordinate system, in a given open interval, how many pairs can meet each other?

Input

There are several test cases in the input.

Each test case begin with one integer N (1 ≤ N ≤ 50000), indicating the number of different lines.
Then two floating numbers L, R follow (-10000.00 ≤ L < R ≤ 10000.00), indicating the interval (L, R).
Then N lines follow, each line contains four floating numbers x1, y1, x2, y2 (-10000.00 ≤ x1, y1, x2, y2 ≤ 10000.00), indicating two different points on the line. You can assume no two lines are the same one.

The input terminates by end of file marker.

Output

For each test case, output one integer, indicating pairs of intersected lines in the open interval, i.e. their intersection point’s x-axis is in (l, r).

Sample Input

  
  
   
   3
0.0 1.0
0.0 0.0 1.0 1.0
0.0 2.0 1.0 2.0
0.0 2.5 2.5 0.0

Sample Output

Author

iSea @ WHU

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

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题意：输入n条直线，和区间(l,r)(注意不含l和r)。问x在(l,r)内时，总共有多少个交点。
题解：先学怎么判断交点。对于两条直线line1和line2,他们在横坐标l和r位置上纵坐标分别为l1,r1,l2,r2.
则(l1-l2)*(r1-r2)<0时有交点。
通过上面的方法我们可以想到逆序数的解法。
先将所有直线根据l递增排序，之后编号1~n，再根据r递减排序，得到一个编号序列。例如3412，其中r3>r4>r1>r2,
又l1<l2<r3<r4,所以3和4有交点，1和2有交点。这符合逆序数的关系：一个数的逆序数是在它之前比他大的数的个数，
当然这里是小的数，原因是为了方便树状数组处理。所以只要根据上述方法排序再求逆序数即可。
不过有些特殊情况需要处理：
1.与y轴平行得线，只要这样的线在(l,r)范围内，则必定跟别的不平行线相交。所以计算下个数在乘积就OK。
2.l和r相同的情况。当l相同时，r递增排序；当r相同时，l递减排序。就能使在l和r上的交点不计算在内。
死活不知道我注释的那求逆序对的写法哪错了，先收藏一下吧
ac代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct s
{
	double a,b;
	int num;
}c[50010];
int cmp1(s x,s y)
{
	if(x.a==y.a)
		return x.b<y.b;
	return x.a<y.a;
}
int cmp2(s x,s y)
{
	if(x.b==y.b)
		return x.a>y.a;
	return x.b>y.b;
}
int d[50010],t;
int low(int x)
{
	return x&(-x);
}
void add(int p,int q)
{
	while(p<=t)
	{
		d[p]+=q;
		p+=low(p);
	}
}
int sum(int p)
{
	int ans=0;
	while(p>0)
	{
		ans+=d[p];
		p-=low(p);
	}
	return ans;
}
int cot[100010];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		double l,r;
		int i,tt=0,ans=0;
		t=0;
		scanf("%lf%lf",&l,&r);
		for(i=0;i<n;i++)
		{
			double x1,x2,y1,y2,k,bn;
			scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
			if(x1==x2)
			{
				if(x1<r&&x1>l)
					tt++;
				continue;
			}
			k=(y1-y2)/(x1-x2);
			bn=y1-k*x1;
			c[t].a=k*l+bn;
			c[t++].b=k*r+bn;
		}
		sort(c,c+t,cmp1);
		for(i=0;i<t;i++)
		{
			c[i].num=i+1;
		}
		sort(c,c+t,cmp2);
		memset(d,0,sizeof(d));
		memset(cot,0,sizeof(cot));
		/*for(i=0;i<t;i++)
		{
			cot[c[i].num]=i+1;
		}
		for(i=1;i<=t;i++)
		{
			add(cot[i],1);
			ans+=i-sum(cot[i]);
		}*/
		for(i=0;i<t;i++)
		{
			add(c[i].num,1);
			ans+=sum(c[i].num-1);
		}
		printf("%dn",ans+tt*t);
	}
}