【LOJ】#3085. 「GXOI / GZOI2019」特技飞行

LOJ#3085. 「GXOI / GZOI2019」特技飞行

这显然是两道题，求\(C\)是一个曼哈顿转切比雪夫后的线段树扫描线

求\(AB\)，对向交换最大化和擦身而过最大化一定分别为最大值和最小值

对向交换最大化是每个点都对向交换

擦身而过最大化需要对向交换最小化，我们一次对向交换相当于交换任意两个数，所以就是每个置换圈的点数-1累加即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,K,tot;
int64 A,B,C,xst,xed;
int64 Min,Max;
int64 y[MAXN][2];
struct Point {
    db x,y;
    Point(db _x = 0.0,db _y = 0.0) {
	x = _x;y = _y;
    }
}P[MAXN * 10];
void getinsect(int a,int b) {
    db k1 = 1.0 * (y[a][1] - y[a][0]) / (xed - xst);
    db k2 = 1.0 * (y[b][1] - y[b][0]) / (xed - xst);
    db t1 = y[a][0] - xst * k1;
    db t2 = y[b][0] - xst * k2;
    db X = (t2 - t1) / (k1 - k2);
    db Y = k1 * X + t1;
    P[++tot] = Point(X + Y,X - Y);
}
namespace task1 {
    int id[MAXN],pos[MAXN],fa[MAXN],siz[MAXN];
    int64 all;
    set<pii > S;
    int getfa(int u) {
	return fa[u] == u ? u : fa[u] = getfa(fa[u]);
    }
    void Main() {
	read(N);read(A);read(B);read(C);read(xst);read(xed);
	for(int i = 1 ; i <= N ; ++i) read(y[i][0]);
	for(int i = 1 ; i <= N ; ++i) {read(y[i][1]);id[i] = i;}
	sort(id + 1,id + N + 1,[](int a,int b){return y[a][1] < y[b][1];});
	for(int i = 1 ; i <= N ; ++i) pos[id[i]] = i;
	for(int i = N ; i >= 1 ; --i) {
	    for(auto t : S) {
		if(t.fi < pos[i]) {++all;getinsect(t.se,i);}
		else break;
	    }
	    S.insert(mp(pos[i],i));
	}
	for(int i = 1 ; i <= N ; ++i) fa[i] = i;
	for(int i = 1 ; i <= N ; ++i) {
	    if(getfa(pos[i]) != getfa(i)) fa[getfa(pos[i])] = getfa(i);
	}
	for(int i = 1 ; i <= N ; ++i) {
	    siz[getfa(i)]++;
	}
	int64 t = 0;
	for(int i = 1 ; i <= N ; ++i) {
	    if(getfa(i) == i) {
		t += siz[i] - 1;
	    }
	}
	Min = all * A;
	Max = t * A + (all - t) * B;
	if(Min > Max) swap(Min,Max);
    }
};
namespace task2 {
    struct qry_node {
	int64 l,r;db y;
	int on;
    }qry[MAXN * 30];
    struct node {
	int l,r;
	int cnt;
    }tr[MAXN * 120];
    int d,all;
    int64 ans,val[MAXN * 30];
    void build(int u,int l,int r) {
	tr[u].l = l;tr[u].r = r;
	tr[u].cnt = 0;
	if(l == r) return;
	int mid = (l + r) >> 1;
	build(u << 1,l,mid);
	build(u << 1 | 1,mid + 1,r);
    }
    void Add(int u,int l,int r,int v) {
	if(tr[u].l == l &&tr[u].r == r) {tr[u].cnt += v;return;}
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(r <= mid) Add(u << 1,l,r,v);
	else if(l > mid) Add(u << 1 | 1,l,r,v);
	else {Add(u << 1,l,mid,v);Add( u << 1 | 1,mid + 1,r,v);}
    }
    bool Query(int u,int p) {
	if(tr[u].cnt) return true;
	if(tr[u].l == tr[u].r) return false;
	int mid = (tr[u].l + tr[u].r) >> 1;
	if(p <= mid) return Query(u << 1,p);
	else return Query(u << 1 | 1,p);
    }
    void Main() {
	read(K);
	int64 p,q,r;
	for(int i = 1 ; i <= K ; ++i) {
	    read(p);read(q);read(r);
	    int64 a = p + q,b = p - q;
	    qry[++all] = (qry_node){a - r,a + r,(db)b - r,1};
	    qry[++all] = (qry_node){a - r,a + r,b + r + 1e-12,-1};
	    val[++d] = a + r;val[++d] = a - r;
	}
	sort(val + 1,val + d + 1);
	d = unique(val + 1,val + d + 1) - val - 1;
	build(1,1,d);
	for(int i = 1 ; i <= tot ; ++i) {
	    qry[++all] = (qry_node){0,i,P[i].y,0};
	}
	sort(qry + 1,qry + all + 1,[](qry_node a,qry_node b){return a.y < b.y || (a.y == b.y && a.on > b.on) ;});
	for(int i = 1 ; i <= all ; ++i) {
	    if(qry[i].on == 0) {
		int t = lower_bound(val + 1,val + d + 1,(int64)ceil(P[qry[i].r].x)) - val;
		if(t <= d) ans += Query(1,t);
	    }
	    else {
		int l = lower_bound(val + 1,val + d + 1,qry[i].l) - val;
		int r = lower_bound(val + 1,val + d + 1,qry[i].r) - val;
		++l;
		Add(1,l,r,qry[i].on);

	    }
	}
	Min += ans * C;
	Max += ans * C;
    }
};
void Solve() {
    task1::Main();
    task2::Main();
    out(Min);space;out(Max);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}