Submission #22790

#	Time	Username	Problem	Language	Result	Execution time	Memory
22790		크리콘 B번 문제는 그리디로 풀려요 (#40)	Donut-shaped Enclosure (KRIII5_DE)	C++98	2 / 7	1500 ms	116900 KiB

DE

#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;

typedef long long lld;
const lld MINF = -2000000000;

struct intv {
	lld x, yl, yr; int sco;
	intv(lld x_=0, lld yl_=0, lld yr_=0, int sco_=0){ x=x_, yl=yl_, yr=yr_, sco=sco_; }
	bool operator< (const intv& c) const {
		if(x == c.x)return sco < c.sco;
		return x<c.x;
	}
} ba[404000];

struct iix {
	int sum, mx;
	iix(int sum_=0, int mx_=0) : sum(sum_), mx(mx_) {}
};

struct segtr {
	int l, r; lld s, e;
	iix key;
	segtr(lld s_=0, lld e_=0) : s(s_), e(e_) {l=NULL, r=NULL;}
} itr[3000000];

int N, bcn, dap; int icn;
lld L, R;

void add(int it, lld mi, lld mx, int v){
	if(mi > mx)return;
	lld l = itr[it].s, r = itr[it].e;
	lld m = ((l-MINF)+(r-MINF))/2 + MINF;
	if(m == r && l < m)m--;
//	printf("%lld %lld %lld / %lld %lld\n", l, m, r, mi, mx);
	if(l == -1 && r == -1 && mi == 0)exit(0);
	if(l == mi && r == mx){
		itr[it].key.sum += v;
		itr[it].key.mx += v;
		return;
	}
	if(itr[it].l == 0)itr[it].l = icn++, itr[itr[it].l] = segtr(l, m);
	if(itr[it].r == 0)itr[it].r = icn++, itr[itr[it].r] = segtr(m+1, r);

	if(mx <= m) add(itr[it].l, mi, mx, v);
	else if(mi > m) add(itr[it].r, mi, mx, v);
	else{
		add(itr[it].l, mi, m, v);
		add(itr[it].r, m+1, mx, v);
	}
	itr[it].key.mx = max(itr[itr[it].l].key.mx, itr[itr[it].r].key.mx) + itr[it].key.sum;
}

lld ys[1204040]; int ycn;

int main(){
	icn++;
	itr[0].s = 0, itr[0].e = 1300000;
	scanf("%d %lld %lld", &N, &L, &R); L--;
	for(int i=0; i<N; i++){
		lld x, y; int s;
		scanf("%lld%lld%d", &x, &y, &s);
		ba[bcn++] = intv(x-R, y-R, y+R, s);
		ba[bcn++] = intv(x-L, y-L, y+L, -s);
		ba[bcn++] = intv(x+L+1, y-L, y+L, s);
		ba[bcn++] = intv(x+R+1, y-R, y+R, -s);
		ys[ycn++] = y-L; ys[ycn++] = y+L;
		ys[ycn++] = y-L-1; ys[ycn++] = y+L-1;
		ys[ycn++] = y-L+1; ys[ycn++] = y+L+1;
		ys[ycn++] = y-R; ys[ycn++] = y+R;
		ys[ycn++] = y-R-1; ys[ycn++] = y+R-1;
		ys[ycn++] = y-R+1; ys[ycn++] = y+R+1;
	}
	sort(ba, ba+bcn);
	sort(ys, ys+ycn); ycn = unique(ys, ys+ycn) - ys;
	for(int i=0; i<bcn; i++){
		int yl = lower_bound(ys, ys+ycn, ba[i].yl) - ys;
		int yr = lower_bound(ys, ys+ycn, ba[i].yr) - ys;
		add(0, yl, yr, ba[i].sco);
		int mx = itr[0].key.mx;
		if(dap < mx)dap = mx;
	}
	printf("%d", dap);
	return 0;
}

Compilation message (stderr)

DE.cpp: In constructor 'segtr::segtr(lld, lld)':
DE.cpp:26:45: warning: converting to non-pointer type 'int' from NULL [-Wconversion-null]
  segtr(lld s_=0, lld e_=0) : s(s_), e(e_) {l=NULL, r=NULL;}
                                             ^
DE.cpp:26:53: warning: converting to non-pointer type 'int' from NULL [-Wconversion-null]
  segtr(lld s_=0, lld e_=0) : s(s_), e(e_) {l=NULL, r=NULL;}
                                                     ^
DE.cpp: In function 'int main()':
DE.cpp:61:35: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d %lld %lld", &N, &L, &R); L--;
                                   ^
DE.cpp:64:34: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%lld%lld%d", &x, &y, &s);
                                  ^

• Subtask 1: 2 / 2
• Subtask 2: 0 / 5