Poj2482 -- Stars in Your Window (Scan Line)

Poj2482 -- Stars in Your Window (Scan Line)

 

Give the coordinates of n stars, each of which has a brightness, and give the length and width of a rectangle. Ask the maximum brightness and (excluding the border) of the stars that the rectangle can contain ).

Assuming that each star is the leftmost bottom vertex of a rectangle, each star can get a rectangle (x, y)-> (x, y, x + w, y + h ), the two high edges of the rectangle are the brightness k and-k of the star. For different rectangles, if the overlap occurs, that is to say, the two stars can appear in the same rectangle, scanning the line to find the maximum possible brightness and.

Note: Because the border is not included, you should first scan the limit with a negative value and count the maximum value only for the limit with a positive value.

 

#include 
 
  #include 
  
   #include using namespace std ;#define LL __int64#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2,r,rt<<1|1#define root 1,num_y,1#define int_rt int l,int r,int rtstruct node{    LL x , y1 , y2 ;    LL k ;}p[30000];LL y[30000] ;int cnt , num_y ;struct node1{    LL max1 , l , r , lazy ;}cl[200000];int cmp(node a,node b) {    return a.x < b.x || ( a.x == b.x && a.k < 0 ) ;}void push_up(int rt) {    cl[rt].max1 = max(cl[rt<<1].max1,cl[rt<<1|1].max1) + cl[rt].lazy;}void create(int_rt) {    cl[rt].max1 = cl[rt].lazy = 0 ;    cl[rt].l = y[l] , cl[rt].r = y[r] ;    if( r - l == 1 ) {        return ;    }    create(lson) ;    create(rson) ;    push_up(rt) ;}void update(LL ll,LL rr,LL k,int rt) {    if( cl[rt].l >= ll && cl[rt].r <= rr ) {        cl[rt].lazy += k ;        cl[rt].max1 += k ;        return ;    }    if( ll < cl[rt<<1].r )        update(ll,min(rr,cl[rt<<1].r),k,rt<<1) ;    if( rr > cl[rt<<1|1].l )        update(max(cl[rt<<1|1].l,ll),rr,k,rt<<1|1) ;    push_up(rt) ;}int main() {    LL n , w , h , xx , yy , k , max1 ;    int i , j ;    while( scanf(%I64d %I64d %I64d, &n, &w, &h) != EOF ) {        cnt = 0 , num_y = 1 ;        max1 = 0 ;        for(i = 1 ; i <= n ; i++) {            scanf(%I64d %I64d %I64d, &xx, &yy, &k) ;            p[cnt].x = xx ; p[cnt].y1 = yy ; p[cnt].y2 = yy+h ;            p[cnt++].k = k ;            p[cnt].x = xx+w ; p[cnt].y1 = yy ; p[cnt].y2 = yy+h ;            p[cnt++].k = -k ;            y[num_y++] = yy ;            y[num_y++] = yy+h ;        }        sort(y+1,y+num_y) ;        num_y = unique(y+1,y+num_y) - (y+1) ;        sort(p,p+cnt,cmp) ;        create(root) ;        for(i = 0 ; i < cnt ; i++) {            update(p[i].y1,p[i].y2,p[i].k,1) ;            if( p[i].k  > 0 )                max1 = max(max1,cl[1].max1) ;        }        printf(%I64d, max1) ;    }    return 0 ;}