• bzoj1069: [SCOI2007]最大土地面积


    传送门

    旋转卡壳。

    答案一定出现在对踵点中。

    //Achen
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<vector>
    #include<cstdio>
    #include<queue>
    #include<cmath>
    const int N=2007; 
    #define For(i,a,b) for(int i=(a);i<=(b);i++)
    #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
    #define eps 1e-15
    typedef long long LL;
    typedef double db;
    using namespace std;
    int n,top;
    db ans;
    
    template<typename T>void read(T &x)  {
        char ch=getchar(); x=0; T f=1;
        while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
        if(ch=='-') f=-1,ch=getchar();
        for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
    }
    
    struct pt {
        db x,y;
        pt(){}
        pt(db x,db y):x(x),y(y){}
    }p[N],q[N];
    
    int dcmp(db x) { return fabs(x)<=eps?0:(x>0?1:-1); }
    bool operator == (const pt&A,const pt&B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0; }
    bool operator < (const pt&A,const pt&B) { return dcmp(A.x-B.x)==0?dcmp(A.y-B.y)<0:dcmp(A.x-B.x)<0; }
    pt operator + (const pt&A,const pt&B) { return pt(A.x+B.x,A.y+B.y); }
    pt operator - (const pt&A,const pt&B) { return pt(A.x-B.x,A.y-B.y); }
    pt operator * (const pt&A,db &B) { return pt(A.x*B,A.y*B); }
    pt operator / (const pt&A,db &B) { return pt(A.x/B,A.y/B); }
    db cross(pt A,pt B) { return A.x*B.y-A.y*B.x; }
    db dot(pt A,pt B) { return A.x*B.x+A.y*B.y; }
    db length(pt A) { return dot(A,A); }
    
    bool cmp(const pt&A,const pt&B) {
        return cross(A-p[1],B-p[1])==0?length(A-p[1])<length(B-p[1]):cross(A-p[1],B-p[1])>0;
    }
    
    void get_ham() {
        For(i,2,n) if(p[i]<p[1]) swap(p[i],p[1]); 
        sort(p+2,p+n+1,cmp);
        q[top=1]=p[1]; q[++top]=p[2];
        For(i,3,n) {
            while(top>1&&dcmp(cross(q[top]-q[top-1],p[i]-q[top-1]))<=0) top--;
            q[++top]=p[i];
        }
    }
    
    void RC() {
        q[top+1]=q[1];
        int now1=2;
        For(i,1,top) {
            while(dcmp(cross(q[i+1]-q[i],q[now1]-q[i])-cross(q[i+1]-q[i],q[now1+1]-q[i]))<0) {
                now1++; if(now1==top+1) now1=1;
            }
            int now2=i+2>top?(i+2)%top:i+2;
            for(int j=(i==top)?1:i+1;j!=now1;(j+1<=top)?j++:j=1) {
                while(dcmp(cross(q[j+1]-q[j],q[now2]-q[j])-cross(q[j+1]-q[j],q[now2+1]-q[j]))<0) {
                    now2++; if(now2==top+1) now2=1;
                }
                ans=max(ans,cross(q[j]-q[i],q[now1]-q[i])+cross(q[now1]-q[i],q[now2]-q[i]));
            }
        }
    }
    
    int main() {
    #ifdef DEBUG
        freopen(".in","r",stdin);
        freopen(".out","w",stdout);
    #endif
        read(n);
        For(i,1,n) { scanf("%lf%lf",&p[i].x,&p[i].y); }
        get_ham(); RC();
        printf("%.3lf
    ",ans/2.0);
        return 0;
    }
    /*
    5
    0 0
    1 0
    1 1
    0 1
    0.5 0.5
    */
    View Code
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  • 原文地址:https://www.cnblogs.com/Achenchen/p/8604405.html
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