HDU5862-Counting Intersections

Counting Intersections

                                                                  Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                            Total Submission(s): 1290    Accepted Submission(s): 402

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

  

   2
4
1 0 1 3
2 0 2 3
0 1 3 1
0 2 3 2
4
0 0 2 0
3 0 3 2
3 3 1 3
0 3 0 2
  

 

Sample Output

  

   4
0
  

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

 

Recommend

wange2014

 

题意：给你n条平行于x轴或平行于y轴的线段，求有多少个交点

解题思路：离散化+扫描线，将所有竖线只保存点坐标，每条竖线下面的点val为1，上面的点val为-1，对所有点和横线按照y轴进行排序，然后从下往上扫即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

struct node
{
    int x1,y,x2;
    friend bool operator<(node a,node b)
    {
        return a.y<b.y;
    }
}a[100009];

struct point
{
    int x,y,val;
    friend bool operator<(point a,point b)
    {
        if(a.y!=b.y) return a.y<b.y;
        else return a.val>b.val;
    }
}p[200009];

int n,cnt,res1,res2;
int x[200009],sum[200009];

int lowbit(int k) {return k&-k;}
void update(int k,int val) {for(;k<cnt;k+=lowbit(k)) sum[k]+=val;}
int getsum(int k) {int ans=0;for(;k;k-=lowbit(k)) ans+=sum[k];return ans;}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cnt=res1=res2=1;
        for(int i=1;i<=n;i++)
        {
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            x[cnt++]=x1,x[cnt++]=x2;
            if(x1==x2)
            {
                if(y1>y2) swap(y1,y2);
                p[res1].x=x1,p[res1].y=y1,p[res1++].val=1;
                p[res1].x=x1,p[res1].y=y2,p[res1++].val=-1;
            }
            else
            {
                if(x1>x2) swap(x1,x2);
                a[res2].y=y1,a[res2].x1=x1,a[res2++].x2=x2;
            }
        }
        sort(x+1,x+cnt);
        cnt=unique(x+1,x+cnt)-x;
        sort(p+1,p+res1);
        sort(a+1,a+res2);
        memset(sum,0,sizeof sum);
        LL ans=0;
        int j=1;
        for(int i=1;i<res2;i++)
        {
            while(j<res1&&(p[j].y<a[i].y||(p[j].y==a[i].y&&p[j].val==1)))
            {
                int k=lower_bound(x+1,x+cnt,p[j].x)-x;
                update(k,p[j].val);
                j++;
            }
            int k=lower_bound(x+1,x+cnt,a[i].x2)-x;
            int kk=lower_bound(x+1,x+cnt,a[i].x1)-x-1;
            ans=ans+1LL*getsum(k)-1LL*getsum(kk);
        }
        printf("%lld\n",ans);
    }
    return 0;
}