HDU 4007 Dave (line tree + discretization + scan line)

Dave Time limit:2000/1000 MS (java/others) Memory limit:65768/65768 K (java/others)
Total submission (s): 2514 accepted Submission (s): 854


Problem Description Recently, Dave is boring, so he often walks around. He finds this some places are too crowded, for example, the ground. He couldn ' t help to the disasters happening recently. Crowded place isn't safe. He knows there are N (1<=n<=1000) people on the ground. Now it wants to know how many people'll be being in a square with the length of R (1<=r<=1000000000). (including boundary).
Input the input contains several cases. For each case there are two positive integers n and R, and then n lines follow. Each gives the (x, y) (1<=x, y<=1000000000) coordinates of people.

Output output The largest number of people in a square with the length of R.
Sample Input

3 2 1 1 2 2 3 3
Sample Output

3 Hint If Two people stand in one place, they are.
Source the 36th ACM/ICPC Asia regional Dalian Site--online Contest
Recommend LCY

The maximum number of squares that are edges r that are parallel to the XY axis

Solving: In fact, can n square double cycle ... is to ask for rectangular intersection ... But for a day a tree of exercise, decisive segment tree discretization scan line together ... 1 A and it's the second ... The mood that cheerful to ... Feel like doing ACM is such a thing is the most fun and happy ~ ~ ~ ~

#include <stdio.h> #include <string.h> #include <stdlib.h> struct edge{int x,y1,y2,flag;}
SEG[2003];
int hasy[2003],tree[8003],lazy[8003],n,all;
    int cmp (const void *a,const void *b) {return * (int *) a>* (int *) b?1:-1;} int cmp2 (const void *a,const void *b) {
    struct Edge c=* (struct edge *) A;
    struct Edge d=* (struct edge *) b;
Return c.x>d.x?1:-1;
    } void down (int pos) {tree[pos<<1]+=lazy[pos];
    tree[pos<<1|1]+=lazy[pos];
    lazy[pos<<1]+=lazy[pos];
    lazy[pos<<1|1]+=lazy[pos];
lazy[pos]=0;
} void up (int pos) {tree[pos]=tree[pos<<1]>tree[pos<<1|1]?tree[pos<<1]:tree[pos<<1|1];}

    void Updata (int l,int r,int pos,int templ,int tempr,int val) {int mid= (L+R) >>1;
        if (TEMPL&LT;=L&AMP;&AMP;R&LT;=TEMPR) {tree[pos]+=val;
        Lazy[pos]+=val;
    Return
    } down (POS);
    if (tempr<=mid) Updata (l,mid,pos<<1,templ,tempr,val); else if (Templ>mid) Updata (mid+1,r,pos<<1|1,templ,tempr,val);
        else {updata (l,mid,pos<<1,templ,mid,val);
    Updata (Mid+1,r,pos<<1|1,mid+1,tempr,val);
Up (POS);

    int mysearch (int x) {int l=0,r=all-1,mid;
        while (l<=r) {mid= (l+r) >>1;
        if (hasy[mid]==x) return mid;
        else if (hasy[mid]>x) r=mid-1;
    else l=mid+1;
} return-1;

    int main () {int i,r,x,y,res; while (scanf ("%d%d", &n,&r) >0) {for (i=0;i<n;i++) {scanf ("%d%d", &x,&y
            );    Seg[2*i].x=x;
            seg[2*i+1].x=x+r+1;
            Hasy[2*i]=seg[2*i].y1=seg[2*i+1].y1=y;
            hasy[2*i+1]=seg[2*i].y2=seg[2*i+1].y2=y+r+1;   seg[2*i].flag=1;
        Seg[2*i+1].flag=-1;
        } qsort (Hasy,2*n,sizeof (hasy[0]), CMP);
        Qsort (Seg,2*n,sizeof (seg[0]), CMP2);
            for (n*=2,all=i=1;i<n;i++) {if (hasy[i]==hasy[i-1]) continue; HaSy[all++]=hasy[i];
        } memset (Tree,0,sizeof (tree));
        memset (lazy,0,sizeof (lazy));
            for (res=i=0;i<n;i++) {x=mysearch (seg[i].y1);
            Y=mysearch (SEG[I].Y2);
            Updata (0,all-1,1,x,y,seg[i].flag);
        if (res<tree[1]) res=tree[1];
    printf ("%d\n", res);
return 0;
 }