[POI 2001+2014acm上海邀请赛]Gold Mine/Beam Cannon 线段树+扫描线

Description 

Byteman, one of the most deserving employee of The Goldmine of Byteland, is about to retire by the end of the year. The Goldmine management would like to reward him in acknowledgment of his conscientious work. As a reward Byteman may receive a small lot - a rectangular piece of the mine‘s area with sides of length s and w parallel to the axes of the coordinate system. He may choose the location of the lot. Of course, a value of the lot depends on the location. The value of the lot is a number of gold nuggets in the area of the lot (if a nugget lies on the border of the lot, then it is in the area of the lot). Your task is to write a program which computes the maximal possible value of the lot (the value of the lot with the best location). In order to simplify we assume that the terrain of the Goldmine is boundless, but the area of gold nuggets occurrence is limited. 

Task

Write a program which:

1.reads the location of gold nuggets, 

2.computes the maximal value of a lot (i.e. the maximal number of gold nuggets on the lot of given size), 

3.writes the result . 

Input 

In the first line there are two positive integers s and w separated by a single space, (1<=s,w<=10 000); they denote the lengths of lot‘s sides - parallel to the OX-axe and OY-axe respectively. There is one positive integer n written in the second line, (1<=n<=15 000). It denotes the number of nuggets in the area of the Goldmine. In the following n lines there are written the coordinates of the nuggets. Each of these lines consists of two integers x and y, (-30 000<=x,y<=30 000), separated by a single space and denoting the x and the y coordinate of a nugget respectively.

Output 

The out should contain exactly one integer equal to the value of the most valuable lot of a given size.

Sample Input 

1 2

12

0 0

1 1

2 2

3 3

4 5

5 5

4 2

1 4

0 5

5 0

2 3

3 2

Sample Output 

4

#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <cstring>
using namespace std;
struct node{
    int t,pos,st;
}ha[200000];
int x1,x2,v;
int _max;
int n,w,h;
int cmp(node x,node y)
{
    return (x.t<y.t);//按照时间先后顺序排序，先删除点再添加点
}
int maxv[400000],addv[400000];
void maintain(int o,int L,int R)
{
    int lc=o<<1;
    int rc=lc+1;
    maxv[o]=0;
    if (R>L)
    {
        maxv[o]=max(maxv[lc],maxv[rc]);
    }
    maxv[o]+=addv[o];
}
void update(int o,int L,int R)
{
    int lc=o<<1;
    int rc=lc+1;
    if (x1<=L&&x2>=R) addv[o]+=v;
    else {
        int M=L+(R-L)/2;
        if (x1<=M) update(lc,L,M);
        if (x2>M) update(rc,M+1,R);
    }
    maintain(o,L,R);
}
void query(int o,int L,int R,int add)
{
    int lc=o<<1;
    int rc=lc+1;
    if (x1<=L&&x2>=R) _max=max(_max,maxv[o]+add);
    else {
        int M=L+(R-L)/2;
        if (x1<=M) query(lc,L,M,addv[o]+add);
        if (x2>M) query(rc,M+1,R,addv[o]+add);
    }
}
int main()
{
    while(~scanf("%d%d%d",&w,&h,&n))
    {
        memset(maxv,0,sizeof(maxv));
        memset(addv,0,sizeof(addv));
        _max=0;
        int o,p,pt=0;
        memset(ha,0,sizeof(ha));
        for (int i=1;i<=n;i++)
            {
                scanf("%d%d",&o,&p);
                ha[++pt].t=o;
                ha[pt].pos=p+30001;
                ha[pt].st=1;
                ha[++pt].t=o+w+1;
                ha[pt].pos=p+30001;
                ha[pt].st=-1;
            }
        sort(ha+1,ha+pt+1,cmp);//将所有时间点排序
        for (int i=1;i<=pt;i++)
        {
            x1=ha[i].pos;
            x2=ha[i].pos+h;
            x2=min(x2,60001);
            v=ha[i].st;
            update(1,1,60001);//更新点数
            if (ha[i].t!=ha[i+1].t||i==pt)//直至同一时间点的所有更新完成之后才输出
                {
                 x1=1;
                 x2=60001;
               <span style="white-space:pre">	</span> query(1,1,60001,0);
            <span style="white-space:pre">	</span>}
        }
        printf("%d\n",_max);
    }
}