初中生数学题

题意

平面上最开始只包含3个点，然后还会依次出现N个点。每新增一个点，请你求出包含这些点的周长最小的多边形的面积。

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思路

采用平衡树维护，每次插入寻找前驱和后继，然后添加节点即可。

注意本题有可能会出现三点共线的情况，可以通过随机指定初始点来解决。

精度略有毒瘤之处。

代码

#include <bits/stdc++.h>
#include <cmath>

using namespace std;

namespace StandardIO {

    template<typename T> inline void read (T &x) {
        x=0;T f=1;char c=getchar();
        for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
        for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
        x*=f;
    }
    template<typename T> inline void write (T x) {
        if (x<0) putchar('-'),x=-x;
        if (x>=10) write(x/10);
        putchar(x%10+'0');
    }

}

using namespace StandardIO;

namespace Solve {
    
    const int N=100100;
    
    int n;
    long long ans;
    struct node {
        long long x,y;
        long double rad;
        node () {}
        node (int _x,int _y):x(_x),y(_y){}
        node (int _x,int _y,long double _r):x(_x),y(_y),rad(_r){}
        node operator - (const node &a) {
            return node(x-a.x,y-a.y);
        }
        node operator + (const node &a) {
            return node(x+a.x,y+a.y);
        } 
        long long operator * (const node &a) {
            return (x*a.y-y*a.x);
        }
    } a[N],S;
    long double X,Y;
    set<node> sexy;
    
    inline bool operator < (const node &x,const node &b) {
        return x.rad<b.rad;
    }
    inline long double rad (node x) {
        return (long double)atan2((x.y-S.y),(x.x-S.x));
    }
    inline node pre (node x) {
        if (sexy.count(x)) return x;
        set<node>::iterator it=sexy.lower_bound(x);
        if (it==sexy.begin()) it=sexy.end();
        return *(--it);
    }
    inline node suc (node x) {
        set<node>::iterator it=sexy.upper_bound(x);
        if (it==sexy.end()) it=sexy.begin();
        return *it;
    }
    inline long long Abs (long long a) {
        return (a>0)?a:-a;
    }
    inline long long calc (node x,node b) {
        return Abs(x*b);
    }
    inline void insert (node x) {
        if (sexy.size()<=2) {
            sexy.insert(x);
            return;
        }
        node l=pre(x),r=suc(x);
        if ((x-l)*(r-l)<=0) return;
        ans+=calc(x-l,r-l);
        while (1) {
            node tmp=pre(x);
            sexy.erase(tmp),l=pre(x);
            if ((x-l)*(l-tmp)>=0) {
                sexy.insert(tmp);
                break;
            }
            ans+=calc(tmp-x,l-x);
        }
        while (1) {
            node tmp=suc(x);
            sexy.erase(tmp),r=suc(x);
            if ((x-r)*(r-tmp)<=0) {
                sexy.insert(tmp);
                break;
            }
            ans+=calc(tmp-x,r-x);
        }
        sexy.insert(x);
    }
    inline void init () {
        long double J=(long double)1.92,Z=(long double)6.08,M=(long double)1.7;
        S=node((long double)((long double)a[1].x*J+(long double)a[2].x*Z+(long double)a[3].x*M)/(long double)(J+Z+M),(long double)((long double)a[1].y*J+(long double)a[2].y*Z+(long double)a[3].y*M)/(long double)(J+Z+M));
        insert(node(a[1].x,a[1].y,rad(a[1])));
        insert(node(a[2].x,a[2].y,rad(a[2])));
        insert(node(a[3].x,a[3].y,rad(a[3])));
        ans+=calc(a[1]-a[3],a[3]-a[2]);
    }

    inline void MAIN () {
        for (register int i=1; i<=3; ++i) read(a[i].x),read(a[i].y);
        init();
        read(n);
        for (register int i=4; i<=n+3; ++i) {
            read(a[i].x),read(a[i].y);
            insert(node(a[i].x,a[i].y,rad(a[i])));
            write(ans),putchar('\n');
        }
    }
    
    #undef int
}

int main () {
    Solve::MAIN();
}