题目链接

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

题意：

给你n(1<=n<=100000)条线段，每个线段平行于坐标轴的x轴或者y轴。问其中相交的点有多少。(线段之间最多只有一个交点。)

题解：

现在我们枚举平行于y轴的直线，然后扫一遍，找出有多少个平行于x轴的与之相交。
对于怎么求有多少个平行于x轴的与之相交。
首先因为我们对于直线的处理是按照x轴坐标排序，为什么这么排序，这样的话对于我们每一次扫描到平行于y轴的直线我们都不需要考虑其他左端点大于该平行于y轴的直线了。
至于查询，由于该平行于y轴的直线只会可能于左端点小于等于他的平行于x轴的直线有关，使用树状数组存储这个值。
至于这个值怎么存储？这里使用的是左端点值+1，右端点下一个点值-1。为什么这么做，首先如果这个点是大于右端点那么在树状数组累计的时候+1-1了。如果在左右端点之间那么数据直接+1了。这样可以方便的统计。

代码：

#include 
      
       using namespace std;const int maxn = 101000;#define lowbit(x) (x&(-x))struct Node{    int type,x,y,y1;    bool operator < (const Node & R)const{        return (x == R.x ? type < R.type : x < R.x);    }}node[maxn*2];int Maxn;int cy[maxn*2];int bi[maxn*2];void add(int add,int n){    for (int i = add; i <= Maxn; i += lowbit(i))        bi[i] += n;}int sum(int n){    int ret = 0;        for (int i = n; i > 0; i -= lowbit(i))        ret += bi[i];    return ret;}map 
       
        mp;int main(){    int t;    scanf("%d",&t);    while (t--){        mp.clear();        memset(bi,0,sizeof bi);        int n;        scanf("%d",&n);        int cnode,ccy;        cnode = ccy = 0;                int x1,x2,y1,y2;        for (int i = 1; i <= n; i++){            scanf("%d %d %d %d",&x1,&y1,&x2,&y2);            if (x1 == x2){                if (y1 > y2)    swap(y1,y2);                    node[++cnode]={1,x1,y1,y2};                cy[++ccy] = y1;                cy[++ccy] = y2;            }else {                if (x1 > x2)    swap(x1,x2);                node[++cnode]={0,x1,y1,1};                node[++cnode]={0,x2+1,y2,-1};                cy[++ccy] = y1;            }        }        sort(cy+1,cy+ccy+1);        int acl = 0;        for (int i = 1; i <= ccy; i++){            if (!mp[cy[i]]) mp[cy[i]] = ++ acl;            }        Maxn = acl;        sort(node+1,node+cnode+1);        long long ans = 0;        for (int i = 1; i <= cnode; i++){            if (node[i].type){                ans += (sum(mp[node[i].y1]) - sum(mp[node[i].y]-1));            }else {                add(mp[node[i].y],node[i].y1);                }            }        printf("%lld\n",ans);    }    return 0;    }