【BZOJ】4056: [Ctsc2015]shallot

題意

在線、可持久化地維護一條二維平面上的折線，支持查詢與任意一條直線的交點個數。
點的個數和操做個數小於\(10^5\)

分析

一條折線能夠用一個序列表示，可持久化序列考慮用可持久化treap。
如何判斷交點？若是有交點，那麼必定與包含這個折線的矩陣有交點。

題解

因此咱們可持久化treap一下便可，雖然這個複雜度很不靠譜，純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;
}