hdu-5862 Counting Intersections(线段树+扫描线)

题目链接：

Counting Intersections

Time Limit: 12000/6000 MS (Java/Others)  

  Memory Limit: 65536/65536 K (Java/Others)

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

题意：

 

求这些与坐标轴平行的线段的交点有多少个;

 

思路：

 

差不多就是原来的一道CF的原题,传送门

交点的个数就是每个线段覆盖的点数的和减去所有线段覆盖的点,而所有的线段覆盖的点数就可以用扫描线算法变成求面积;

 

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+4;
int n,x1,x2,y3,y2,rec[2*N],num;
struct no
{
    int l,r,h,flag;
};
no line[8*N];
struct nod
{
    int l,r,cover;
    ll sum;
};
nod tree[8*N];
int cmp(no x,no y)
{
    return x.h<y.h;
}
void build(int node,int L,int R)
{
    tree[node].l=L,tree[node].r=R;
    tree[node].cover=tree[node].sum=0;
    if(L>=R)return ;
    int mid=(L+R)>>1;
    build(2*node,L,mid);
    build(2*node+1,mid+1,R);
}
void Pushup(int node)
{
    if(tree[node].cover)
    {
        tree[node].sum=rec[tree[node].r+1]-rec[tree[node].l];
    }
    else
    {
        if(tree[node].l==tree[node].r)tree[node].sum=0;
        else tree[node].sum=tree[2*node].sum+tree[2*node+1].sum;
    }
}
void update(int node,int L,int R,int x)
{
    if(L<=tree[node].l&&R>=tree[node].r)
    {
        tree[node].cover+=x;
        Pushup(node);
        return ;
    }
    int mid=(tree[node].l+tree[node].r)>>1;
    if(L>mid) update(2*node+1,L,R,x);
    else if(R<=mid)update(2*node,L,R,x);
    else
    {
        update(2*node,L,mid,x);
        update(2*node+1,mid+1,R,x);
    }
    Pushup(node);
}
int bi(int x)
{
    int L=1,R=num-1,mid;
    while(L<=R)
    {
        mid=(L+R)>>1;
        if(rec[mid]==x)return mid;
        else if(rec[mid]>x)R=mid-1;
        else L=mid+1;
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d",&n);
    ll sum=0;
    int cnt=1;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d%d",&x1,&y3,&x2,&y2);
            if(x1>x2)swap(x1,x2);
            if(y3>y2)swap(y3,y2);
            rec[cnt] = line[cnt].l = x1;
            line[cnt].r = x2+1;
            line[cnt].h = y3;
            line[cnt++].flag = 1;
            line[cnt].l = x1;
            rec[cnt] = line[cnt].r = x2+1;
            line[cnt].h = y2+1;
            line[cnt++].flag = -1;
            if(x1==x2)sum=sum+abs(y2-y3)+1;
            else sum=sum+abs(x1-x2)+1;
    }
    sort(line+1,line+cnt,cmp);
    sort(rec+1,rec+cnt);
    num = 2;
    for(int i = 2;i < cnt;i++)
    {
        if(rec[i]!=rec[i-1])rec[num++]=rec[i];
    }
    build(1,1,num-1);
    ll ans=0;
    for(int i = 1;i < cnt-1;i++)
    {
        int fx = bi(line[i].l);
        int fy = bi(line[i].r)-1;
        if(fx <= fy)
        {
            update(1,fx,fy,line[i].flag);
        }
        ans+=tree[1].sum*(ll)(line[i+1].h-line[i].h);
    }
   // cout<<ans<<endl;
    //cout<<sum<<" ";
    printf("%lld\n",sum-ans);
}
    return 0;
}