hdu 5091 Beam Cannon（線段樹+掃描線+離散化）

Beam Cannon

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 457    Accepted Submission(s): 175

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

 

題意：在一個平面內有N個人，用一個W*H的矩形去圍這些人（邊上的也算）， 求最大人數。

思路：以x從小到大排序，y值離散化，投影到y軸上，那麼對於每個人的縱座標，y，y+h就是

每個星星可以影響到的矩形 然後x，x+w+1就是一個進入事件和一個出去事件，其所帶的值互

爲相反數. node[1].val 保存當前的最大值 當所有的矩形都遍歷一遍 取其中的最大值就是ans。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=20005;

struct node
{
    int x,y1,y2,val;
    void fun(int xx,int yy1,int yy2,int v)
    {
        x=xx,y1=yy1,y2=yy2;
        val=v;
    }
}b[maxn];
struct tree
{
    int l,r,add,val;
}a[maxn*4];

int n,w,h,cnt,Y[maxn];

bool cmp(node p,node q)
{
    return p.x<q.x;
}

void build(int l,int r,int k)
{
    a[k].l=l,a[k].r=r;
    a[k].add=a[k].val=0;
    if(l==r)  return ;
    int mid=(l+r)/2;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
}

void pushdown(int k)
{
    a[2*k].val+=a[k].add;
    a[2*k].add+=a[k].add;
    a[2*k+1].val+=a[k].add;
    a[2*k+1].add+=a[k].add;
    a[k].add=0;
}

void insert(int l,int r,int c,int k)
{
    if(Y[a[k].l]==l && Y[a[k].r]==r)
    {
         a[k].val+=c;
         a[k].add+=c;
    }
    else
    {
         pushdown(k);
         int mid=(a[k].l+a[k].r)/2;
         if(Y[mid]>=r)       insert(l,r,c,2*k);
         else if(Y[mid]<l)   insert(l,r,c,2*k+1);
         else
         {
              insert(l,Y[mid],c,2*k);
              insert(Y[mid+1],r,c,2*k+1);
         }
         a[k].val=max(a[2*k].val,a[2*k+1].val);
    }
}

void input()
{
    cnt=0;
    int x,y;
    scanf("%d %d",&w,&h);
    for(int i=0;i<n;i++)
    {
        scanf("%d %d",&x,&y);
        y+=20000;
        Y[cnt]=y;
        b[cnt++].fun(x,y,y+h,1);
        Y[cnt]=y+h;
        b[cnt++].fun(x+w+1,y,y+h,-1);
    }
    sort(Y,Y+cnt);
    sort(b,b+cnt,cmp);
    cnt=unique(Y,Y+cnt)-Y;
}

void solve()
{
    build(0,cnt-1,1);
    int ans=0;
    for(int i=0;i<2*n;i++)
    {
        insert(b[i].y1,b[i].y2,b[i].val,1);
        ans=max(ans,a[1].val);
    }
    printf("%d\n",ans);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n<0)  break;
        input();
        solve();
    }
    return 0;
}