HDU 5862 Counting Intersections 扫描线+树状数组

题目链接：

http://acm.split.hdu.edu.cn/showproblem.php?pid=5862

Counting Intersections

Time Limit: 12000/6000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others) #### 问题描述 > 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. #### 输入 > 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. #### 输出 > For each test case, output one line, the number of intersection.

样例

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

给你若干个平行于坐标轴的，长度大于0的线段，且任意两个线段没有公共点，不会重合覆盖。问有多少个交点。

题解

扫描线+树状数组
首先把平行于x轴的线段和平行于y轴的线段分开存。对于平行于y轴的线段，我们按它的上端点排降序，对于平行于x轴的线段，我们按照它们的高度排降序。

然后我们遍历一遍平行于x轴的线段，考虑有多少条线段与当前遍历的平行于x轴的线段所在的直线相交的线段，并且更新到树状数组中，那么与当前平行于x轴的线段相交的垂直线段为sum(r)-sum(l-1),l、r分别为当前线段的左右端点。

那么，如何维护与某条直线相交的垂直线段呢？
由于我们遍历平行于x轴的直线是按高度从高到低的，对于上端点高于当前遍历高度的所有垂直线段都压到树状数组中，并且把它们的下端点（大的优先）压到优先队列中，然后再从优先队列中踢掉下端点大于当前遍历高度的，同时更新树状数组（删除操作）。

由于数据范围比较大，横坐标需要离散化。

代码

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) ;//cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define reu(i,a,b) for(int i=a;i<=(b);i++)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;

//start----------------------------------------------------------------------

const int maxn=4e5+10;

LL sumv[maxn];

struct Node{
	int val,u,v;
	Node(int val,int u,int v):val(val),u(u),v(v){}
	bool operator < (const Node& tmp) const {
		return val>tmp.val;
	}
};

bool cmp(const Node& n1,const Node& n2){
	return n1.v>n2.v;
}

LL sum(int x){
	LL ret=0;
	while(x>0){
		ret+=sumv[x];
		x-=(x&-x);
	} 
	return ret;
}

void add(int x,int v){
	while(x<maxn){
		sumv[x]+=v;
		x+=(x&-x);
	}
}

VI ha; 
vector<Node> col,row;
int n;

void init(){
	ha.clear();
	col.clear();
	row.clear();
	clr(sumv,0);
}

int main() {
	int tc;
	scanf("%d",&tc);
	while(tc--){
		
		scanf("%d",&n);
		init();
		
		rep(i,0,n){
			int u1,v1,u2,v2;
			scanf("%d%d%d%d",&u1,&v1,&u2,&v2);
			if(u1==u2){
				if(v1>v2) swap(v1,v2);
				col.push_back(Node(u1,v1,v2));
			}else{
				if(u1>u2) swap(u1,u2);
				row.push_back(Node(v1,u1,u2));
			}
			ha.push_back(u1);
			ha.push_back(u2);
		}
		
		sort(all(ha));
		ha.erase(unique(all(ha)),ha.end());
		
		sort(all(col),cmp);
		sort(all(row));
		
		priority_queue<pair<int,int> > pq;
		int p=0;
		LL ans=0;
		rep(i,0,row.sz()){
			int hi=row[i].val;
			bug(hi);
			while(p<col.sz()&&col[p].v>=hi){
				int id=lower_bound(all(ha),col[p].val)-ha.begin()+1;
				pq.push(mkp(col[p].u,id));
				add(id,1);
				p++;
			}
			while(!pq.empty()&&pq.top().X>hi){
				add(pq.top().Y,-1);
				pq.pop();
			}
			int l=lower_bound(all(ha),row[i].u)-ha.begin()+1;
			int r=lower_bound(all(ha),row[i].v)-ha.begin()+1;
			ans+=sum(r)-sum(l-1); 
		}
		
		printf("%lld\n",ans);
	}
	return 0;
}

//end-----------------------------------------------------------------------

/*
4
1 0 1 1
0 1 2 1
3 1 3 2
1 2 1 3
*/