HDU 5862 Counting Intersections（树状数组+扫描线+离散化） Counting Intersections

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我是靠谱客的博主包容路人，最近开发中收集的这篇文章主要介绍HDU 5862 Counting Intersections（树状数组+扫描线+离散化） Counting Intersections，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Counting Intersections

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1296 Accepted Submission(s): 407

Problem Description

Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection.

The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments with length 0.

Input

The first line contains an integer T, indicates the number of test case.

The first line of each test case contains a number n(1<=n<=100000), the number of segments. Next n lines, each with for integers, x1, y1, x2, y2, means the two endpoints of a segment. The absolute value of the coordinate is no larger than 1e9.

Output

For each test case, output one line, the number of intersection.

Sample Input

Sample Output

4
0

Author

BUPT

Source

2016 Multi-University Training Contest 10

题意：

给你几根与坐标轴平行的线，问你产生几个交点。

POINT：

树状数组的区间值是保存了y区间内有几根横线。

把y轴离散化。横线只要看成2个点，和竖线一起依x轴从小到大排序。

然后从左向右扫描，扫描到左端点，就y点+1，右端点就-1，扫描到竖线，就查询竖线上的y区间内右几根横线。（单点修改，区间查询）

注意端点与线重合的情况，优先扫描左端点，然后再扫描竖线，最后是右端点，不然答案会少。画一个右端点和竖线重合的图思考一下。

（就是为了去补这题多校，才去做了很多线段树和树状数组，终于把这题1A了，思考时间也不长，写的很快还没有错，那么线段树和树状数组也就告一段落，下面学习RMQ吧。自己还有很多要学的呢。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 100050*5;
#define  LL long long
struct node//离散纵坐标
{
    int x,y1,y2;
    int flag;//零代表竖线， 1代表横线左端点 -1右端点
}dian[N];
int haha[N];
int mm;
int yy[N];
int num[N];
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int c)
{
    while(x<=mm)
    {
        num[x]+=c;
        x+=lowbit(x);
    }
}
LL query(int x)
{
    LL ans=0;
    while(x>=1)
    {
        ans+=(LL)num[x];
        x-=lowbit(x);
    }
    return ans;
}
int Discretization(int y)
{
    return (int)(lower_bound(yy+1,yy+1+mm,y)-yy);
}
bool cmd(node a,node b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    else
        return a.flag>b.flag;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(dian,0,sizeof dian);
        memset(haha,0,sizeof haha);
        memset(yy,0,sizeof yy);
        memset(num,0,sizeof num);
        int m=0;
        int my=0;
        for(int i=1;i<=n;i++)
        {
            int x1,x2,y1,y2;
            scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
            if(x1>x2) swap(x1,x2);
            if(y1>y2) swap(y1,y2);
            if(x1==x2)
            {
                dian[++m].x=x1;
                dian[m].flag=0,dian[m].y1=y1,dian[m].y2=y2;
                yy[++my]=y1;
                yy[++my]=y2;
            }
            else
            {
                dian[++m].x=x1;
                dian[m].flag=1,dian[m].y1=dian[m].y2=y1;
                dian[++m].x=x2;
                dian[m].flag=-1,dian[m].y1=dian[m].y2=y2;
                yy[++my]=y1;
            }
        }
        sort(yy+1,yy+1+my);
        mm=1;
        for(int i=2;i<=my;i++)
        {
            if(yy[i-1]!=yy[i]) yy[++mm]=yy[i];
        }
        sort(dian+1,dian+1+m,cmd);
        LL ans=0;
        for(int i=1;i<=m;i++)
        {
            if(dian[i].flag==0)//竖线
            {
                int Y=Discretization(dian[i].y2);
                int y=Discretization(dian[i].y1);
                ans+=query(Y)-query(y-1);
            }
            else if(dian[i].flag==-1)//右端点
            {
                int y=Discretization(dian[i].y1);
                add(y,-1);
            }
            else
            {
                int y=Discretization(dian[i].y1);
                add(y,1);
            }
        }
        printf("%lldn",ans);
    }

}