Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 201 Accepted Submission(s): 77
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
Recommend
zhouzeyong
题意：
给一个开区间（l,r），给n个直线，问这些直线在这段区间里面有多少个交点。‘’
思路：
如果暴力求解的话，时间复杂度为n^2，大概为10^11,肯定会超时。
由于题目将x的坐标限定在了一个区间（l,r）内，考虑如下情况：
设直线1分别与x=l，x=r相交与L1,R1;直线2分别相交于L2，R2。若两直线在此区间内相交，必有
L1<L2&&R1>R2 或者 L1>L2&&R1<R2
（想象若两直线在（l,r）相交，则经过交点后，纵坐标的大小顺序一定会改变）
此即为逆序数对的性质，于是题目转化为了求逆序数对的个数：
还需注意两点：
另外很重要的一点，数据为实型数，需要对数据进行离散化处理，如此才能应用树状数组
关于离散化：http://blog.csdn.net/gokou_ruri/article/details/7723378
代码：
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) (x&(-x))
const int N=5e4+10;
struct node{
double a,b;
int num;
};
node e[N];
int c[N];
double l,r;
int cmp1(node x,node y) //left 递增排序
{
if(x.a==y.a)
return x.b<y.b;
return x.a<y.a;
}
int cmp2(node x,node y) //right 递减排序
{
if(x.b==y.b)
return x.a>y.a;
return x.b>y.b;
}
void add(int x,int val)
{
while(x<N)
{
c[x]+=val;
x+=lowbit(x);
}
}
int sum(int x)
{
int ans=0;
while(x>0)
{
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
int n;
int t,tt,i,j,ans;
double x1,y1,x2,y2,k,b;
while(scanf("%d",&n)!=EOF)
{
t=tt=0;
ans=0;
scanf("%lf%lf",&l,&r);
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);//两个点的横纵坐标
if(x1==x2)//横坐标相等 表示竖直的线 平行与y轴
{
if(l<x1&&x1<r)
tt++;
continue;
}
k=(y2-y1)/(x2-x1);//y=kx+b
b=y1-k*x1;
e[t].a=l*k+b;
e[t++].b=r*k+b;
}
//******************//
//离散化,由于想树状数组中add的是left，所以只对left离散化即可
sort(e,e+t,cmp1);
for(i=0;i<t;i++)//编号
e[i].num=i+1;//树状数组无下标0
//*******************//
sort(e,e+t,cmp2);//递减排序 3412
memset(c,0,sizeof(c));
for(i=0;i<t;i++)//统计
{
add(e[i].num,1);
ans+=sum(e[i].num-1);
}
printf("%d
",ans+tt*t);//加上平行y轴的直线所产生的交点
}
}