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  • 计算几何 模板

    凸包

    POJ 1113

    #include <cstdio>
    #include <cstdlib>
    #include <iostream>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    
    using namespace std;
    typedef long long ll;
    
    const double eps = 1e-9;
    const double PI = acos(-1.0);
    
    inline int sgn(double x) {
    	if (x < -eps) return -1;
    	if (x > eps) return 1;
    	return 0;
    }
    
    struct vec {
    	double x, y;
    	vec() {x = y = 0;}
    	vec(double _x, double _y): x(_x), y(_y) {}
    
    	vec operator + (vec v) { return vec(x + v.x, y + v.y);}
    	vec operator - (vec v) { return vec(x - v.x, y - v.y);}
    	vec operator * (double v) {return vec(x * v, y * v);}
    	vec operator / (double v) {return vec(x / v, y / v);}
    
    	double operator * (vec v) {return x * v.x + y * v.y;}
    	double len() {return hypot(x, y);}
    	double lensqr() {return x * x + y * y;}
    } V[1005], Z[1005];
    
    double cross(vec a, vec b) {
    	return a.x * b.y - a.y * b.x;
    }
    
    double dist(double x1, double y1, double x2, double y2) {
    	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
    }
    
    bool cmpXY(vec a, vec b) {
    	if (sgn(a.x - b.x)) return a.x < b.x;
    	return a.y < b.y;
    }
    
    double convex_hull(vec* v, int n, vec *z) {
    	sort(v, v + n, cmpXY);
    	int m = 0;
    	for (int i = 0; i < n; i++) {
    		while (m > 1 && cross(z[m - 1] - z[m - 2], v[i] - z[m - 2]) <= 0) --m;
    		z[m++] = v[i];
    	}
    	int k = m;
    	for (int i = n - 2; i >= 0; i--) {
    		while (m > k && cross(z[m - 1] - z[m - 2], v[i] - z[m - 2]) <= 0) --m;
    		z[m++] = v[i];
    	}
    	if (n > 1) --m;
    	double ret = 0;
    	for (int i = 0; i < m; i++)
    		ret += dist(z[i].x, z[i].y, z[(i + 1) % m].x, z[(i + 1) % m].y);
    	return ret;
    }
    
    int main() {
    	//freopen("in.txt","r",stdin);
    	int n;
    	double r;
    	while (~scanf("%d", &n)) {
    		scanf("%lf", &r);
    		for (int i = 0; i < n; i++)
    			scanf("%lf%lf", &V[i].x, &V[i].y);
    		double ret = convex_hull(V, n, Z);
    		ret += PI * r * 2.0;
    		int ans = ret + 0.5;
    		printf("%d
    ", ans);
    	}
    	return 0;
    }
    

    BZOJ 1185

    #include<cstdio>
    #include<cmath>
    #include<ctime>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<queue>
    #include<set>
    #define eps 1e-8
    #define inf 1000000000
    using namespace std;
    double ans=1e60;
    int n,top;
    struct P{
        double x,y;
        P(){}
        P(double _x,double _y):x(_x),y(_y){}
        friend bool operator<(P a,P b){
            return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y;
        }
        friend bool operator==(P a,P b){
            return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;
        }
        friend bool operator!=(P a,P b){
            return !(a==b);
        }
        friend P operator+(P a,P b){
            return P(a.x+b.x,a.y+b.y);
        }
        friend P operator-(P a,P b){
            return P(a.x-b.x,a.y-b.y);
        }
        friend double operator*(P a,P b){
            return a.x*b.y-a.y*b.x;
        }
        friend P operator*(P a,double b){
            return P(a.x*b,a.y*b);
        }
        friend double operator/(P a,P b){
            return a.x*b.x+a.y*b.y;
        }
        friend double dis(P a){
            return sqrt(a.x*a.x+a.y*a.y);
        }
    }p[50005],q[50005],t[5];
    bool cmp(P a,P b)
    {
        double t=(a-p[1])*(b-p[1]);
        if(fabs(t)<eps)return dis(p[1]-a)-dis(p[1]-b)<0;
        return t>0;
    }
    void graham()
    {
        for(int i=2;i<=n;i++)
            if(p[i]<p[1])
                swap(p[i],p[1]);
        sort(p+2,p+n+1,cmp);
        q[++top]=p[1];
        for(int i=2;i<=n;i++)
        {
            while(top>1&&(q[top]-q[top-1])*(p[i]-q[top])<eps)top--;
            q[++top]=p[i];
        }
        q[0]=q[top];
    }
    void RC()
    {
        int l=1,r=1,p=1;
        double L,R,D,H;
        for(int i=0;i<top;i++)
        {
            D=dis(q[i]-q[i+1]);
            while((q[i+1]-q[i])*(q[p+1]-q[i])-(q[i+1]-q[i])*(q[p]-q[i])>-eps)p=(p+1)%top;
            while((q[i+1]-q[i])/(q[r+1]-q[i])-(q[i+1]-q[i])/(q[r]-q[i])>-eps)r=(r+1)%top;
            if(i==0)l=r;
            while((q[i+1]-q[i])/(q[l+1]-q[i])-(q[i+1]-q[i])/(q[l]-q[i])<eps)l=(l+1)%top;
            L=(q[i+1]-q[i])/(q[l]-q[i])/D,R=(q[i+1]-q[i])/(q[r]-q[i])/D;
            H=(q[i+1]-q[i])*(q[p]-q[i])/D;
            if(H<0)H=-H;
            double tmp=(R-L)*H;
            if(tmp<ans)
            {
                ans=tmp;
                t[0]=q[i]+(q[i+1]-q[i])*(R/D);
                t[1]=t[0]+(q[r]-t[0])*(H/dis(t[0]-q[r]));
                t[2]=t[1]-(t[0]-q[i])*((R-L)/dis(q[i]-t[0]));
                t[3]=t[2]-(t[1]-t[0]);
            }
        }
    }
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        graham();
        RC();
        printf("%.5lf
    ",ans);
        int fir=0;
        for(int i=1;i<=3;i++)
            if(t[i]<t[fir])
                fir=i;
        for(int i=0;i<=3;i++)
            printf("%.5lf %.5lf
    ",t[(i+fir)%4].x,t[(i+fir)%4].y);
        return 0;
    }
    

    最大空凸包

    POJ 1259 HDU 6219

    dp[i][j]表示以<i,j>为以o为原点选的最后一个三角形的边。(i>j)
    需要确定三角形内无点
    那么dp[i][j]=dp[j][k]+S(i,j,o)。需要<j,i>与<k,j>形成凸边。

    #include <cstdio>
    #include <cstdlib>
    #include <iostream>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    
    using namespace std;
    typedef long long ll;
    
    const double eps = 1e-9;
    const double PI = acos(-1.0);
    const int maxn = 405;
    double dp[maxn][maxn];
    int T,n;
    
    inline int sgn(double x) {
        if (x < -eps) return -1;
        if (x > eps) return 1;
        return 0;
    }
    
    struct vec {
        double x, y;
        vec() {x = y = 0;}
        vec(double _x, double _y): x(_x), y(_y) {}
    
        vec operator + (vec v) { return vec(x + v.x, y + v.y);}
        vec operator - (vec v) { return vec(x - v.x, y - v.y);}
        vec operator * (double v) {return vec(x * v, y * v);}
        vec operator / (double v) {return vec(x / v, y / v);}
    
        double operator * (vec v) {return x * v.x + y * v.y;}
        double len() {return hypot(x, y);}
        double lensqr() {return x * x + y * y;}
    } V[maxn], Z[maxn],o;
    
    double cross(vec a, vec b) {
        return a.x * b.y - a.y * b.x;
    }
    
    double dist(double x1, double y1, double x2, double y2) {
        return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
    }
    
    bool cmpangle(vec a,vec b){
    	int k=sgn(cross(a-o,b-o));
    	if(k==0){
    		return (a-o).lensqr()<(b-o).lensqr();
    		//return dist(a.x,a.y,o.x,o.y)<dist(b.x,b.y,o.x,o.y);
    	} else return k>0;
    }
    
    double calc_empty_convex(vec Z[],int n){
    	for(int i=0;i<=n;i++)
    		for(int j=0;j<=n;j++)
    			dp[i][j]=0;
    	double ret=0;
    	int j,k;
    	for(int i=0;i<n;i++){
    		j=i-1;
    		while(j>=0 && sgn(cross(Z[i]-o,Z[j]-o))==0) j--;
    		bool flg=(j==i-1);
    		while(j>=0){
    			k=j-1;
    			while(k>=0&&cross(Z[i]-Z[j],Z[j]-Z[k])>0) k--;
    			double area=fabs(cross(Z[i]-o,Z[j]-o))*0.5;
    			if(k>=0) area+=dp[j][k];
    			// cout<<area<<endl;
    			if(flg) dp[i][j]=area;
    			ret=max(area,ret);
    			j=k;
    		}
    		if(flg){
    			for(int c=1;c<i;c++)
    				dp[i][c]=max(dp[i][c],dp[i][c-1]);
    		}
    	}
    	//cout<<ret<<endl;
    	return ret;
    }
    
    double empty_convex(vec V[],int n){
    	double ret=0;
    	for(int i=0;i<n;i++){
    		o=V[i];
    		int top=0;
    		for(int j=0;j<n;j++)
    			if(V[j].y>o.y||(sgn(V[j].y-o.y)==0&&sgn(V[j].x-o.x)>=0)){
    				Z[top++]=V[j];
    			}
    		sort(Z,Z+top,cmpangle);
    		ret=max(ret,calc_empty_convex(Z,top));
    	}
    	return ret;
    }
    
    int main() {
        //freopen("in.txt","r",stdin);
        cin>>T;
        while(T--){
        	scanf("%d",&n);
        	for(int i=0;i<n;i++)
        		scanf("%lf%lf",&V[i].x,&V[i].y);
        	double res = empty_convex(V,n);
        	printf("%.1f
    ",res);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/foreignbill/p/7806709.html
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