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[APIO2018] Circle selection 选圆圈(假题解)


目录


题面

自己去

LOJ

上找

Sol

直接排序然后

KDTree

查询
然后发现

TLE

然后把点旋转一下,就过了。。

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

const int maxn(3e5 + 5);
const double eps(1e-3);
const double alpha(acos(-1) / 5);
const double cosa(cos(alpha));
const double sina(sin(alpha));

int n, ans[maxn], op, rt;

struct Data{
    double d[2], r;
    int id;

    IL int operator <(RG Data b) const{
        return d[op] < b.d[op];
    }
} a[maxn];

IL int Cmp(RG Data x, RG Data y){
    return x.r != y.r ? x.r > y.r : x.id < y.id;
}

struct KDTree{
    double d[2], mn[2], mx[2], r;
    int ls, rs, id;
} tr[maxn];

IL void Chkmax(RG double &x, RG double y){
    if(y > x) x = y;
}

IL void Chkmin(RG double &x, RG double y){
    if(y < x) x = y;
}

IL void Update(RG int x){
    tr[x].mn[0] = tr[x].mx[0] = tr[x].d[0];
    tr[x].mn[1] = tr[x].mx[1] = tr[x].d[1];
    RG int ls = tr[x].ls, rs = tr[x].rs;
    if(ls){
        Chkmin(tr[x].mn[0], tr[ls].mn[0]), Chkmin(tr[x].mn[1], tr[ls].mn[1]);
        Chkmax(tr[x].mx[0], tr[ls].mx[0]), Chkmax(tr[x].mx[1], tr[ls].mx[1]);
    }
    if(rs){
        Chkmin(tr[x].mn[0], tr[rs].mn[0]), Chkmin(tr[x].mn[1], tr[rs].mn[1]);
        Chkmax(tr[x].mx[0], tr[rs].mx[0]), Chkmax(tr[x].mx[1], tr[rs].mx[1]);
    }
}

IL int Build(RG int l, RG int r, RG int nop){
    RG int x = (l + r) >> 1; op = nop;
    nth_element(a + l, a + x, a + r + 1);
    tr[x].d[0] = a[x].d[0], tr[x].d[1] = a[x].d[1];
    tr[x].id = a[x].id, tr[x].r = a[x].r;
    if(l < x) tr[x].ls = Build(l, x - 1, nop ^ 1);
    if(r > x) tr[x].rs = Build(x + 1, r, nop ^ 1);
    Update(x);
    return x;
}

# define Sqr(x) ((x) * (x))

IL int Check(RG int x, RG Data p){
    RG double x1 = tr[x].d[0], y1 = tr[x].d[1], x2 = p.d[0], y2 = p.d[1], r1 = tr[x].r, r2 = p.r;
    return Sqr(x1 - x2) + Sqr(y1 - y2) - eps <= Sqr(r1 + r2);
}

IL int Cut(RG int x, RG Data p){
    RG double x2 = p.d[0], y2 = p.d[1], r = p.r + p.r;
    if(x2 + r + eps < tr[x].mn[0]) return 1;
    if(y2 + r + eps < tr[x].mn[1]) return 1;
    if(x2 - r > tr[x].mx[0] + eps) return 1;
    if(y2 - r > tr[x].mx[1] + eps) return 1;
    return 0;
}

IL void Modify(RG int x, RG Data p){
    if(Cut(x, p)) return;
    if(!ans[tr[x].id] && Check(x, p)) ans[tr[x].id] = p.id;
    if(tr[x].ls) Modify(tr[x].ls, p);
    if(tr[x].rs) Modify(tr[x].rs, p);
}

int main(){
    n = Input();
    for(RG int i = 1; i <= n; ++i){
        RG double x = Input(), y = Input(), r = Input();
        a[i] = (Data){x * cosa - y * sina, x * sina + y * cosa, r, i};
    }
    rt = Build(1, n, 0), sort(a + 1, a + n + 1, Cmp);
    for(RG int i = 1; i <= n; ++i)
        if(!ans[a[i].id]) Modify(rt, a[i]);
    for(RG int i = 1; i <= n; ++i) printf("%d ", ans[i]);
    return 0;
}

转载自:https://blog.csdn.net/oi_Konnyaku/article/details/80516774