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();
}