HDOJ 题目3465 Life is a Line（树状数组求逆序对，第二种写法）

Life is a Line

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1706    Accepted Submission(s): 389

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

  

   1
  

 

Author

iSea @ WHU

 

Source

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

 

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刚刚写完一篇博客，仔细有想了一下下，瞬间觉悟，知道哪错了

第二种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("%d\n",ans+tt*t);
	}
}