题意
在线、可持久化地维护一条二维平面上的折线，支持查询与任意一条直线的交点个数。
点的个数和操作个数小于\(10^5\)
分析
一条折线可以用一个序列表示，可持久化序列考虑用可持久化treap。
如何判断交点？如果有交点，那么一定与包含这个折线的矩阵有交点。
题解
所以我们可持久化treap一下即可，虽然这个复杂度很不靠谱，纯rp算法。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Lim=20000000, M=1e5+10, oo=~0u>>1;
struct node *null;
struct node {
int x, y, s, mx[2], mn[2];
node *c[2], *ch[2];
void up() {
s=c[0]->s+c[1]->s+1;
mx[0]=max(x, max(c[0]->mx[0], c[1]->mx[0]));
mn[0]=min(x, min(c[0]->mn[0], c[1]->mn[0]));
mx[1]=max(y, max(c[0]->mx[1], c[1]->mx[1]));
mn[1]=min(y, min(c[0]->mn[1], c[1]->mn[1]));
if(c[0]!=null) ch[0]=c[0]->ch[0]; else ch[0]=this;
if(c[1]!=null) ch[1]=c[1]->ch[1]; else ch[1]=this;
}
void init(int _x, int _y) {
x=mx[0]=mn[0]=_x;
y=mx[1]=mn[1]=_y;
c[0]=c[1]=null;
ch[0]=ch[1]=this;
s=1;
}
}Po[Lim], *iT=Po, *t[M];
node *newnode(int x, int y) {
iT->init(x, y);
return iT++;
}
void init() {
null=iT++;
null->init(0, 0);
null->mx[0]=null->mx[1]=-oo;
null->mn[0]=null->mn[1]=oo;
null->s=0;
}
node *build(int l, int r) {
if(l>r) {
return null;
}
node *ret;
int x, y, mid=(l+r)>>1;
node *le=build(l, mid-1);
scanf("%d%d", &x, &y);
node *ri=build(mid+1, r);
ret=newnode(x, y);
ret->c[0]=le;
ret->c[1]=ri;
ret->up();
return ret;
}
node *update(int p, int X, int Y, node *x) {
node *y;
if(x->s==p) {
y=newnode(X, Y);
y->c[0]=x;
y->up();
return y;
}
y=iT++;
*y=*x;
if(y->c[0]->s>=p) {
y->c[0]=update(p, X, Y, y->c[0]);
}
else {
y->c[1]=update(p-y->c[0]->s-1, X, Y, y->c[1]);
}
y->up();
return y;
}
ll cross(int x0, int y0, int x, int y) {
return (ll)x0*y-(ll)x*y0;
}
bool jiao(int x, int y, int xx, int yy, int x0, int y0, int X, int Y) {
ll a=cross(x-x0, y-y0, X, Y),
b=cross(xx-x0, yy-y0, X, Y);
a=a>=0?a>0:-1;
b=b>=0?b>0:-1;
return a*b<=0;
}
bool jiao(node *x, int x0, int y0, int X, int Y) {
return jiao(x->mn[0], x->mx[1], x->mn[0], x->mn[1], x0, y0, X, Y) ||
jiao(x->mn[0], x->mn[1], x->mx[0], x->mn[1], x0, y0, X, Y) ||
jiao(x->mx[0], x->mn[1], x->mx[0], x->mx[1], x0, y0, X, Y) ||
jiao(x->mx[0], x->mx[1], x->mn[0], x->mx[1], x0, y0, X, Y);
}
int query(int x0, int y0, int X, int Y, node *x) {
if(x->s<=1 || !jiao(x, x0, y0, X, Y)) {
return 0;
}
int ret=0;
if(x->c[0]!=null && jiao(x->x, x->y, x->c[0]->ch[1]->x, x->c[0]->ch[1]->y, x0, y0, X, Y)) {
++ret;
}
if(x->c[1]!=null && jiao(x->x, x->y, x->c[1]->ch[0]->x, x->c[1]->ch[0]->y, x0, y0, X, Y)) {
++ret;
}
return query(x0, y0, X, Y, x->c[0])+query(x0, y0, X, Y, x->c[1])+ret;
}
int main() {
init();
int n, m, cn, last=0;
scanf("%d%d%d", &n, &m, &cn);
t[0]=build(1, n);
for(int kk=1; kk<=m; ++kk) {
static char s[5];
int T, x0, y0, x, y;
scanf("%s", s);
if(s[0]=='H') {
scanf("%d%d%d%d%d", &T, &x0, &y0, &x, &y);
if(cn) {
x0^=last;
y0^=last;
x^=last;
y^=last;
}
t[kk]=t[T];
printf("%d\n", last=query(x0, y0, x, y, t[kk]));
}
else {
scanf("%d%d%d%d", &T, &x0, &x, &y);
if(cn) {
x^=last;
y^=last;
}
t[kk]=update(x0, x, y, t[T]);
}
}
return 0;
}