卧薪尝胆,厚积薄发。
HAOI2011 防线修建
Date: Sat Nov 03 15:08:06 CST 2018 In Category: NoCategory

Description:

支持一个数据结构,删除一个点,询问上凸壳的长度。
$1\leqslant n\leqslant 10^5$

Solution:

删除凸包上的点是一个经典的难解问题,所以时光倒流转化为插入,可以用一个 $splay$ 动态维护凸壳并实时维护当前凸壳长度,注意答案的更新有很多细节。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<cstring>
using namespace std;
inline int rd()
{
register int res = 0,f = 1;
register char c = getchar();
while(!isdigit(c))
{
if(c == '-')f = -1;
c = getchar();
}
while(isdigit(c))
{
res = (res << 1) + (res << 3) + c - '0';
c = getchar();
}
return res * f;
}
int n,x,y,m;
#define MAXN 100010
struct point
{
int x,y;
}p[MAXN];
#define MAXQ 200010
int cnt[MAXN];
struct query
{
int k,type;
double res;
}q[MAXQ];
int qn;
double ans = 0;
double dis(point a,point b)
{
return sqrt(double(a.x - b.x) * double(a.x - b.x) + double(a.y - b.y) * double(a.y - b.y));
}
struct node
{
int fa,c[2];
point p;
}t[MAXN];
int root;
int ptr = 0;
int newnode(){return ++ptr;}
int id(int k){return (t[t[k].fa].c[0] == k ? 0 : 1);}
void connect(int k,int f,int p){t[k].fa = f;t[f].c[p] = k;return;}
void rotate(int x)
{
int y = t[x].fa,z = t[y].fa,fx = id(x),fy = id(y);
connect(t[x].c[fx ^ 1],y,fx);
connect(y,x,fx ^ 1);connect(x,z,fy);
return;
}
void splay(int x)
{
while(t[x].fa != 0)
{
int y = t[x].fa;
if(t[y].fa == 0){rotate(x);break;}
if(id(x) == id(y)){rotate(y);rotate(x);}
else{rotate(x);rotate(x);}
}
root = x;
return;
}
int pre()
{
int cur = t[root].c[0];
if(cur == 0)return 0;
while(t[cur].c[1] != 0)cur = t[cur].c[1];
return cur;
}
int nxt()
{
int cur = t[root].c[1];
if(cur == 0)return 0;
while(t[cur].c[0] != 0)cur = t[cur].c[0];
return cur;
}
double slope(point a,point b){return double(a.y - b.y) / double(a.x - b.x);}
void remove(int k)
{
splay(k);
int nrt = nxt();
splay(nrt);
connect(t[k].c[0],nrt,0);
return;
}
void maintain(int k)
{
splay(k);
int pr = pre(),nx = nxt();
if(slope(t[k].p,t[pr].p) <= slope(t[k].p,t[nx].p))
{
ans -= dis(t[k].p,t[pr].p) + dis(t[k].p,t[nx].p);
ans += dis(t[pr].p,t[nx].p);
remove(k);
return;
}
while(true)
{
splay(k);int p = pre();
splay(p);int pp = pre();
if(pp == 0)break;
if(slope(t[pp].p,t[p].p) <= slope(t[p].p,t[k].p))
{
ans -= dis(t[pp].p,t[p].p) + dis(t[p].p,t[k].p);
ans += dis(t[k].p,t[pp].p);
remove(p);
}
else break;
}
while(true)
{
splay(k);int p = nxt();
splay(p);int pp = nxt();
if(pp == 0)break;
if(slope(t[pp].p,t[p].p) >= slope(t[p].p,t[k].p))
{
ans -= dis(t[pp].p,t[p].p) + dis(t[p].p,t[k].p);
ans += dis(t[k].p,t[pp].p);
remove(p);
}
else break;
}
return;
}
void insert(point k)
{
int cur = root;
int pos = newnode();
t[pos].p = k;
while(true)
{
if(k.x < t[cur].p.x)
{
if(t[cur].c[0] != 0)cur = t[cur].c[0];
else
{
splay(cur);
int pr = pre();
splay(pr);splay(cur);
ans -= dis(t[t[cur].c[0]].p,t[cur].p);
ans += dis(t[t[cur].c[0]].p,t[pos].p) + dis(t[pos].p,t[cur].p);
connect(t[cur].c[0],pos,0);connect(pos,cur,0);
maintain(pos);
return;
}
}
else if(k.x > t[cur].p.x)
{
if(t[cur].c[1] != 0)cur = t[cur].c[1];
else
{
splay(cur);
int nx = nxt();
splay(nx);splay(cur);
ans -= dis(t[t[cur].c[1]].p,t[cur].p);
ans += dis(t[t[cur].c[1]].p,t[pos].p) + dis(t[pos].p,t[cur].p);
connect(t[cur].c[1],pos,1);connect(pos,cur,1);
maintain(pos);
return;
}
}
else
{
splay(cur);
if(k.y <= t[cur].p.y)return;
ans -= dis(t[pre()].p,t[cur].p) + dis(t[nxt()].p,t[cur].p);
t[cur].p = k;
ans += dis(t[pre()].p,t[cur].p) + dis(t[nxt()].p,t[cur].p);
maintain(cur);
}
}
return;
}
int main()
{
scanf("%d%d%d",&n,&x,&y);
scanf("%d",&m);
for(int i = 1;i <= m;++i)
{
p[i].x = rd();p[i].y = rd();
}
scanf("%d",&qn);
for(int i = 1;i <= qn;++i)
{
q[i].type = rd();
if(q[i].type == 1)
{
q[i].k = rd();
++cnt[q[i].k];
}
}
ans = dis((point){0,0},(point){x,y}) + dis((point){x,y},(point){n,0});
root = newnode();
t[root].p.x = x;t[root].p.y = y;
int lef = newnode(),rig = newnode();
t[lef].p = (point){0,0};t[rig].p = (point){n,0};
connect(lef,root,0);connect(rig,root,1);
for(int i = 1;i <= m;++i)
{
if(cnt[i] == 0)
{
insert(p[i]);
}
}
for(int i = qn;i >= 1;--i)
{
if(q[i].type == 1)
{
--cnt[q[i].k];
if(cnt[q[i].k] == 0)insert(p[q[i].k]);
}
else
{
q[i].res = ans;
}
}
for(int i = 1;i <= qn;++i)
if(q[i].type == 2)
printf("%.2lf\n",q[i].res);
return 0;
}
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