HDU5862 Counting Intersections

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.

InputThe 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. 
OutputFor 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

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求当前线段与坐标轴平行的直线的交点

可以用扫描线的，扫描线or离散化都是一种很神奇的存在方式

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+;
struct Node
{
    int f,x,y,y1;
    bool operator <(const Node &R)const
    {
        return (x==R.x?f<R.f:x<R.x);
    }
} a[N];
int Maxn;
int yy[N],c[N];
void add(int x,int n)
{
    for(int i=x; i<=Maxn; i+=i&-i)c[i]+=n;
}
int sum(int x)
{
    int ans=;
    for(int i=x; i>; i-=i&-i)ans+=c[i];
    return ans;
}
unordered_map<int,int>M;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        M.clear(),memset(c,,sizeof c);
        int n,ctot=,tot=;
        scanf("%d",&n);
        for(int i=,x1,x2,y1,y2; i<n; i++)
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            if(x1==x2)
            {
                if(y1>y2)swap(y1,y2);
                a[++ctot]= {,x1,y1,y2};
                yy[++tot]=y1;
                yy[++tot]=y2;
            }
            else
            {
                if(x1>x2)swap(x1,x2);
                a[++ctot]= {,x1,y1,};
                a[++ctot]= {,x2+,y2,-};
                yy[++tot]=y1;
            }
        }
        sort(yy+,yy+tot+);
        Maxn=;
        for(int i=; i<=tot; i++)if(!M[yy[i]])M[yy[i]]=++Maxn;
        sort(a+,a+ctot+);
        long long ans=;
        for(int i=; i<=ctot; i++)
        {
            if(a[i].f)ans+=(sum(M[a[i].y1])-sum(M[a[i].y]-));
            else add(M[a[i].y],a[i].y1);
        }
        printf("%lld\n",ans);
    }
    return ;
}