HDU - 5862 Counting Intersections （扫描线应用）

HDU - 5862

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：给你很多条平行于坐标轴的线段，然后求交点的个数。

思路：把平行于x轴的线段看成两个点，进点权值是+1，而出点权值为-1，把平行y轴的线，看作区间查询，注意：

入点和查询在同一个位置，先更新入点，出点和查询在同一个位置，那就先查询。当然首先得离散化y坐标~

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define ll long long
const int maxn = 1e5 + 10;
struct node
{
    int si,sj,pi,pj,w;
    node(){}
    node(int a,int b,int c,int d,int e)
    {si = a; sj = b; pi = c; pj = d; w = e;}
}line[maxn << 1];

int yid[maxn << 1];
int ynum,num,t,n,m;
ll val[maxn << 4];

void init()
{
    num = 0;    ynum = 1;
    memset(val,0,sizeof(val));
}

int getid(int x)
{
    return lower_bound(yid + 1,yid + 1 + ynum , x) - yid;
}

bool cmp(node a,node b)
{
    if(a.si == b.si)
    {
        return a.w > b.w;
    }
    return a.si < b.si;
}
void updata(int id,int l,int r,int u,int v)
{
    //cout << l <<" " << r << " " <>1;
    if(u <= mid) updata(id << 1, l , mid, u, v);
    else updata(id<<1|1, mid + 1, r, u, v);

    val[id] = val[id<<1] + val[id<<1|1];
}
ll query(int id,int l,int r,int ql,int qr)
{
    if(r < ql || l > qr) return 0;
    if(l >= ql && r <= qr) return val[id];

    int mid = l+r>>1;
    ll ans = 0;
    if(ql <= mid) ans += query(id<<1, l , mid, ql , qr);
    if(qr > mid) ans += query(id<<1|1,mid+1,r,ql,qr);
    return ans;
}
int main()
{
    //freopen("D:\\in.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        int anum = 0,bnum = 0;
        for(int i = 0; i < n; i++)
        {
            int a,b,c,d;
            scanf("%d%d%d%d",&a,&b,&c,&d);
            yid[ynum++] = b, yid[ynum++] = d;
            if(b == d)
            {
                anum++;
                if(a > c)   swap(a,c);
                line[num++] = node(a,b,0,0, 1);
                line[num++] = node(c,d,0,0,-1);
            }
            else
            {
                bnum++;
                line[num++] = node(a,b,c,d,0);
            }
        }
        if(anum == 0 || bnum == 0)
        {
            printf("0\n");
            continue;
        }
        sort(line, line + num, cmp);
        //离散化
        sort(yid + 1 , yid + 1 + ynum);
        ynum = unique(yid + 1,yid + 1 + ynum) - yid;
        ll ans = 0;
        for(int i = 0; i < num; i++)
        {
            int w = line[i].w;
            if(w != 0)
            {
                int id = getid(line[i].sj);
                updata(1,1,ynum-1,id,w);
            }
            else
            {
                int l = getid(line[i].sj), r = getid(line[i].pj);
                if(l > r) swap(l,r);
                ll temp = query(1,1,ynum-1,l,r);
                ans += temp;
            }
        }
        cout << ans << endl;
    }
	return 0;
}