hdu 5091 Beam Cannon 离散化+扫描线+线段树

Beam Cannon

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 551    Accepted Submission(s): 207

Problem Description

Recently, the γ galaxies broke out Star Wars. Each planet is warring for resources. In the Star Wars, Planet X is under attack by other planets. Now, a large wave of enemy spaceships is approaching. There is a very large Beam Cannon on the Planet X, and it is very powerful, which can destroy all the spaceships in its attack range in a second. However, it takes a long time to fill the energy of the Beam Cannon after each shot. So, you should make sure each shot can destroy the enemy spaceships as many as possible.

To simplify the problem, the Beam Cannon can shot at any area in the space, and the attack area is rectangular. The rectangle parallels to the coordinate axes and cannot rotate. It can only move horizontally or vertically. The enemy spaceship in the space can be considered as a point projected to the attack plane. If the point is in the rectangular attack area of the Beam Cannon(including border), the spaceship will be destroyed.

 

Input

Input contains multiple test cases. Each test case contains three integers N(1<=N<=10000, the number of enemy spaceships), W(1<=W<=40000, the width of the Beam Cannon’s attack area), H(1<=H<=40000, the height of the Beam Cannon’s attack area) in the first line, and then N lines follow. Each line contains two integers x,y (-20000<=x,y<=20000, the coordinates of an enemy spaceship). 

A test case starting with a negative integer terminates the input and this test case should not to be processed.

 

Output

Output the maximum number of enemy spaceships the Beam Cannon can destroy in a single shot for each case.

 

Sample Input

2 3 4
0 1
1 0
3 1 1
-1 0
0 1
1 0
-1

 

Sample Output

2
2

 

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5091

题意：有n个点，给你w*h的框框，问你最多可以框住几个点，边缘也算。

做法：把每个点x-w/2，y-h/2，  与x+w/2，y+h /2，作一个矩形，可以知道，只有那个框框的中心在这个矩形中就可以覆盖这个点。然后就把所有点的矩形画出来，计算最大重合的层数就行了。实际操作中  可以把每个矩形看作  左下角为 x，y，右上角为x+w，y+h。 也就相当于一起平移。最大重合层数不变。

这题和我之前做得算面积的线段树不同。因为这里关注的不在是面积，所以也就不再关注宽度了。所以这里 线段树里的每个点0-（k-1）代表的不是一段长度的状态了，而是每一个点的状态。

所以本来的 r=Bin(s[i].r,k,x)-1;这一句 由原模版变成了r=Bin(s[i].r,k,x)。

然后就是基本功，把线段树改成算最大值的了。cover记录每个区间最大值。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#define ll __int64
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const ll maxn=50010;

struct Seg
{
	ll l,r,h;
	ll flag;
	Seg(){}
	Seg(ll a,ll b,ll c,ll d):l(a),r(b),h(c),flag(d){}
	bool operator<(const Seg &hh) const 
	{
		return h<hh.h;
	}
};

struct Seg s[maxn<<3];
ll cover[maxn<<2],add[maxn<<2];
ll x[maxn<<2];
ll ans;
void PushUp(ll rt,ll l,ll r)
{
	cover[rt]=max(cover[rt<<1],cover[rt<<1|1])+add[rt];
}

void Update(ll L,ll R,ll f,ll l,ll r,ll rt)
{
	if(L<=l && r<=R)
	{
		cover[rt]+=f;//这个点的覆盖
		add[rt]+=f; 
		return ;
	}
	ll m=(l+r)>>1;
	if(R<=m) Update(L,R,f,lson);
	else if(L>m) Update(L,R,f,rson);
	else
	{
		Update(L,R,f,lson);
		Update(L,R,f,rson);
	}
	PushUp(rt,l,r);
}

ll Bin(ll k,ll n,ll x[])
{
	ll l,r,m;
	l=0,r=n-1;
	while(l<=r)
	{
		m=(l+r)>>1;
		if(x[m]==k) return m;
		else if(x[m]>k) r=m-1;
		else l=m+1;
	}
	return -1;
}
set<ll> ss;
set<ll>::iterator it;
int main()
{
	ll n;
	ll i;
	ll ww,hh;
	
	ll xx,yy;
	while(scanf("%I64d",&n),n!=-1)
	{
		
		scanf("%I64d%I64d",&ww,&hh);
		ll m=0;
		ss.clear();
		for(i=0;i<n;i++)
		{ 
			scanf("%I64d %I64d",&xx,&yy);

			ss.insert(xx+ww);
			ss.insert(xx);

			s[m++]=Seg(xx,xx+ww,yy,1);
			s[m++]=Seg(xx,xx+ww,yy+hh,-1); 
		}
		sort(s,s+m);
		ll k=0;
		for(it=ss.begin();it!=ss.end();it++)
			x[k++]=*it;

		memset(cover,0,sizeof cover);
		memset(add,0,sizeof add);

		ans=0;
		for(i=0;i<m-1;i++)
		{
			ll l=Bin(s[i].l,k,x);
			ll r=Bin(s[i].r,k,x);
			if(l<=r) Update(l,r,s[i].flag,0,k-1,1);
			ans=max(ans,cover[1]); 
		}
		printf("%I64d\n",ans);
	}
	return 0;

}