• bzoj2300#2300. [HAOI2011]防线修建


    题解:带删点的维护凸包,1.删点2.查询凸包周长
    题解:倒着做就成了带加点的维护凸包,加点时维护一下周长就没了

    //#pragma GCC optimize(2)
    //#pragma GCC optimize(3)
    //#pragma GCC optimize(4)
    //#pragma GCC optimize("unroll-loops")
    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define db double
    #define mp make_pair
    #define pb push_back
    #define pi acos(-1.0)
    #define ll long long
    #define vi vector<int>
    #define mod 1000000007
    #define ld long double
    //#define C 0.5772156649
    //#define ls l,m,rt<<1
    //#define rs m+1,r,rt<<1|1
    #define pll pair<ll,ll>
    #define pil pair<int,ll>
    #define pli pair<ll,int>
    #define pii pair<int,int>
    #define ull unsigned long long
    //#define base 1000000000000000000
    #define fin freopen("a.txt","r",stdin)
    #define fout freopen("a.txt","w",stdout)
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
    inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
    template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
    template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
    inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
    inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
    
    using namespace std;
    
    const ull ba=233;
    const db eps=1e-8;
    const ll INF=0x3f3f3f3f3f3f3f3f;
    const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f;
    
    struct node{
        ll x,y;
        node(){}
        node(ld _x,ld _y){x=_x,y=_y;}
        bool operator <(const node&rhs)const{
            return x<rhs.x||(x==rhs.x&&y<rhs.y);
        }
    };
    #define sqr(a) (a)*(a)
    db dis(node a,node b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
    db ans=0;
    bool ok=0;
    struct hull:set<node>{
        bool bad(iterator it)
        {
            if(it==begin()||next(it)==end())return 0;
            return (it->y-prev(it)->y)*(next(it)->x-it->x)<(next(it)->y-it->y)*(it->x-prev(it)->x);
        }
        void update(node x)
        {
            if(find(x)!=end())return ;
            iterator it=insert(x).fi;
            if(bad(it)){erase(it);return ;}
            if(it!=begin())ans+=dis(*it,*prev(it));
            if(next(it)!=end())ans+=dis(*it,*next(it));
            if(it!=begin()&&next(it)!=end())ans-=dis(*prev(it),*next(it));
            while(it!=begin()&&bad(prev(it)))
            {
                if(prev(it)!=begin())
                {
                    ans-=dis(*prev(it),*prev(prev(it)));
                    ans+=dis(*prev(prev(it)),*it);
                }
                ans-=dis(*it,*prev(it));
                erase(prev(it));
            }
            while(next(it)!=end()&&bad(next(it)))
            {
                if(next(next(it))!=end())
                {
                    ans-=dis(*next(it),*next(next(it)));
                    ans+=dis(*next(next(it)),*it);
                }
                ans-=dis(*it,*next(it));
                erase(next(it));
            }
        }
    }h;
    pii p[N];
    node a[N];
    db res[N];
    bool ban[N];
    int main()
    {
        int n,x,y;scanf("%d%d%d",&n,&x,&y);
        h.update(node(0,0));h.update(node(x,y));h.update(node(n,0));
        int m;scanf("%d",&m);
        for(int i=1;i<=m;i++)scanf("%lld%lld",&a[i].x,&a[i].y);
        int q;scanf("%d",&q);
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&p[i].fi);
            if(p[i].fi==1)scanf("%d",&p[i].se),ban[p[i].se]=1;
        }
        for(int i=1;i<=m;i++)if(!ban[i])h.update(a[i]);
    //    printf("%f %f
    ",ans,2.0*sqrt(5));
        for(int i=q;i>=1;i--)
        {
            if(p[i].fi==2)res[i]=ans;
            else h.update(a[p[i].se]);//,printf("%lld %lld--
    ",a[p[i].se].x,a[p[i].se].y);
        }
        for(int i=1;i<=q;i++)if(p[i].fi==2)printf("%.2f
    ",res[i]);
        return 0;
    }
    /********************
    
    ********************/
    
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/10656873.html
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